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Ghent, 10 June 2013
Authors
Steven De Meester, Jo Dewulf (UGent)
Lex Roes, Martin Patel (UU)
Stefanie Hellweg (ETH)
Prospective Sustainability Assessment of Technologies: Development of Basic Engineering modules for prospective estimations of the material flows and energy requirements
Report prepared within the EC 7th framework project
Project no: 227078
Project acronym: PROSUITE
Project title: Development and application of a standardized methodology for the
PROspective SUstaInability assessment of TEchnologies
Start date project: 1 November 2009
Duration: 4 years
Deliverable No: Interim Deliverable 1.2
Date due: Month 24
Date submitted: Month 24 (1st version)
Date submitted: Month 43 (revised)
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For more information on this report please contact:
Ghent University, Department of Sustainable organic Chemistry and Technology, EnVOC
Coupure Links 653
B-9000 Ghent
Belgium
www.envoc.ugent.be
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Executive summary
After determining the goal and scope of a study, the starting point of a sustainability assessment is
to compile reliable data inventory. Especially in prospective assessments, which is the goal of the
PROSUITE project, doing so is not always possible as the technology might not be implemented and
data is not readily available. The goal of this report is therefore to develop basic engineering
modules that allow the estimation of material and energy flows for process flow diagrams of future
technologies. For this purpose, engineering modules were constructed for 25 Basic Unit Operations:
3 reaction types (chemical reactions, incineration for heat and power and fermentation), 7
separation processes (evaporation, distillation, filtration, sedimenting centrifuges, electrostatic
precipitation, electrodialysis and pressure swing adsorption), 9 physical mechanical processes
(mechanical compression – single stage, mechanical compression – multi stage, pumping
incompressible fluids, pumping incompressible fluids through packings, agitation and mixing of
liquids and suspensions, comminuting, fluidization, pneumatic drying and pneumatic conveying,
conveying solids and fans, blowers & vacuum pumps), 3 types of utilities (heating, cooling and
steam generation) and furthermore 3 modules focusing on the processing and use of materials
(electric manufacturing processes, material requirements as a function of mechanical properties
and fuel use of cars). The processes can be assembled to simulate datasets in the foreground
system, obtaining data compatible with life cycle practices, which allows coupling the modelled
system to a background system.
It was attempted to keep these modules user friendly, including the use of default and rule of
thumb values. Furthermore, parameters can be modified, which allows fitting the Basic Unit
Operations easily to the specific application, whilst results are made compatible for LCA studies.
These points are a major advantage in comparison to the use of generic data or other modelling
options.
Of these twenty-five parameterized modules, ten were tested in a case study. Most of these
modules yield relatively good results compared to the real values with an overall average relative
approximation error of 22%. When these modules are used in the case study to estimate the carbon
footprint of a biorefinery, an accuracy of 96% was achieved. It should, however, be noted that there
is a compensation between over- and underestimation in the case studies conducted. Furthermore,
three major shortcomings were identified. Firstly, whereas the inventory of the core unit process
was obtained with an acceptable error, the supporting unit operations such as pumping are very
specific and difficult to obtain whilst not being negligible. Secondly, a factory is complex and
processes are not independent from each other; a successful application of the BUO approach
therefore requires a careful integration of heat use instead of a linear summation of heat use of
different processes. Thirdly the list of selected BUO is still limited to twenty-two, whereas many
other operations are also applied in industry. For example, in biorefineries, operations such as
cyclones and dryers are frequently used. More parameterized modules should thus be developed in
the future.
It is clear that there is no ’silver bullet’ for an accurate and generic estimation of process flow
diagrams. A discrepancy exists between the level of detail of the module and the ease of use and
thus required time and expertise. Each technology has its own features, and the more case specific
data is used, the more accurate the results will be. As results can differ significantly per specific
application, and (small) modifications of the processes are possible in new technologies,
determining statistical uncertainty was not possible. Nevertheless, while more case studies are
necessary for confirmation of the results in this study, the currently available modules produce
relatively reliable results. The possibility of integration of these modules as parameterized unit
operations in LCA software and economic assessment tools can be of major value to obtain a
detailed data inventory necessary for prospective sustainability assessments. The modules should
however be used with care and with a realistic design of a production chain in mind.
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Table of contents
Executive summary ............................................................................................ 3
Table of contents ............................................................................................... 5
Introduction ...................................................................................................... 7
Goal... ........................................................................................................... 10
Part I: Development of the engineering modules ..................................................... 11
1. Principles ............................................................................................... 12
1.1 General ........................................................................................... 12
1.2 Units .............................................................................................. 14
1.3 Time aspect...................................................................................... 14
2. Unit operations included ............................................................................ 15
2.1 Reactions ......................................................................................... 15
2.2 Separation processes ........................................................................... 29
2.3 Physical mechanical processes ............................................................... 46
2.4 Utilities ........................................................................................... 75
2.5 Processing and use of materials .............................................................. 84
Part II: Validation of the engineering modules ....................................................... 102
1. Introduction .......................................................................................... 103
2. The biorefinery case study ....................................................................... 103
3. Results and discussion ............................................................................. 106
3.1 Validation examples ........................................................................... 106
3.2 Overview of the validation................................................................... 107
3.3 Using the BUO approach in LCA ............................................................. 109
General conclusions ........................................................................................ 111
References ................................................................................................... 112
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List of tables:
Table 1: preferred quantitative and qualitative units used in the calculation tool ................................... 14 Table 2: Included setups for energy generation ........................................................................... 21 Table 3: Stoichiometric fermentation reactions and current and future product yields per substrate consumed
(Patel et al., 2006) ............................................................................................................. 25 Table 4: Elemental formula for micro-organisms (Harding, 2008) ....................................................... 26 Table 5: Default steam savings in one to triple effect evaporators ..................................................... 30 Table 6: Typical solids concentrations in the underflow of a centrifugation ........................................... 36 Table 7: typical centrifuge configurations and power (pw) (Perry & Green, 1999) .................................... 38 Table 8: Default adiabatic coefficients for different ranges ............................................................. 46 Table 9: Default evaporator efficiencies for different types of evaporators .......................................... 47 Table 10: Default roughness factors of different piping material ....................................................... 54 Table 11: Equivalent lengths for different piping parts ................................................................... 55 Table 12: Dimensionless loss coefficient for gradual pipe enlargements ............................................... 56 Table 13: Default pump efficiencies for pumping incompressible fluids ............................................... 57 Table 14: Sphericity of different packing particles ........................................................................ 59 Table 15: Power numbers for newtonian fluids in the laminar (KL), intermediate (P0) and turbulent region (KT)
for different types of impellers (McCabe et al., 2004) ................................................................... 62 Table 16: typical rotational speeds for different impellers in mixing vessels (McCabe et al., 2004) ............... 64 Table 17: Default pumping numbers for types of impellers .............................................................. 65 Table 18: Work Index for various materials in kWh/tonne (Perry & Green, 1999) .................................... 67 Table 19: Typical pressure differences covered in vacuum pumps, fans and blowers ................................ 74 Table 20: Default efficiencies ................................................................................................ 74 Table 21: Steam tables ........................................................................................................ 81 Table 22: Energy use of different types of injection molding machines (Thiriez, 2005) ............................. 85 Table 23:Energy use of four different types of milling machines (Dahmus & Gutowski, 2004) ...................... 87 Table 24: Energy use for the production of a 0.13-µm metal microprocessor. The active area on the wafers is
261 cm2 consisting of 1 cm2 dies (Murphy et al., 2003) .................................................................. 88 Table 25: Ductile iron induction melter energy usage per tonne shipped (Jones, 2007) ............................. 88 Table 26: Energy use of two different waterjet machines (Kurd, 2004) ................................................ 88 Table 27: Characteristic values CWR, CWL and CWa for different driving cycles ......................................... 98
List of figures:
Figure 1A (left) : ethanol from corn, available in separation, not available in a database. Figure 1B (right):
ethanol from corn after protein a database (Jungbluth et al., 2007). .................................................. 8 Figure 2: the structure of the basic engineering modules, based on Van Der Vorst et al. (2009) ................... 13 Figure 3: The environmental and economic aspects of the Basic Unit Operations .................................... 13 Figure 4: Example of a gasification at 1000K with an air/fuel ratio of 0.4 and biomass fuel with 90% dry matter
(DM) content .................................................................................................................... 20 Figure 5: Determination of the minimum reflux ratio from the Equilibrium curve and operating line in a
distillation. Point D is the composition of the distillate, whilst point F is the composition of the feed (Perry &
Green, 1999) .................................................................................................................... 32 Figure 6: the heat required per kg ethanol for the ethanol distillation (at 85°C) in MJ/kg ethanol after a
fermentation to obtain 94% ethanol/water ................................................................................ 33 Figure 7: Example of an electrodialysis process (Wikipedia) ............................................................. 41 Figure 8: The kinetic energy correction factor for Herschel-Bulkley fluid foods (Valentas et al., 1997) .......... 51 Figure 9: Power number correlations ........................................................................................ 63 Figure 10: Closed steam system .............................................................................................. 79 Figure 11: Time and speed pattern of the New European Driving Cycle................................................ 97 Figure 12: Willans lines and resulting trend lines of a 1.4 l TSI gasoline engine (90 kW) for low output and low
rpm ............................................................................................................................... 99 Figure 13: Comparison of car efficiency results using the two available algorithms ............................... 100
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Introduction
Technological development is at the basis of the industrial system. Its continuous development,
enhanced by profitability, has resulted in an exponential growth of human population and welfare,
resulting in a severe environmental impact. However, this technological progress will not be stopped
because of the resulting impact. In contrary, it will keep on evolving, but it should start creating
opportunities in a sustainable way. This means that new technologies should be introduced with a
minimal environmental impact and maximal welfare creation. These evolutions can be assessed by
using Life Cycle Assessment (LCA) in addition to production cost analysis and life cycle costing
(LCC). Basically, the LCA methodology analyzes the impact of flows between the natural
environment and the technosphere. Similarly to an economic assessment, this environmental
assessment methodology thus requires a mass and energy balance of processes in the technosphere.
The data inventory is therefore at the basis of sustainability assessment of technologies.
The problem however, is that new technologies should be assessed before large scale
implementation and that the required mass and energy data is often not available. Currently,
assessments are mainly executed with the scope on well implemented technologies based on data
originating from 2 sources:
Existing setups (e.g. an industrial plant)
LCI databases
Often a foreground system is chosen of which data is collected in an existing plant. Such a system
can be defined as those processes of the system that are regarding their selection or mode of
operation directly affected by decisions analyzed in the study (European Commission JRC/IES,
2010). Afterwards the data collected within this system is coupled to a database containing a similar
type of inventory of existing products or services to cover the background system of the production
chain. However, often the assessed foreground system consists of a certain sequence of industrial
processes which is inherent to the existing setup. This type of assessments does not offer the
freedom to modify the sequence or type of Basic Unit Operations (BUO)1 and the input and output
parameters of the production chains which would change in new technologies. For example in the
biorefinery sector, ethanol can be made from corn. This data inventory is gathered and readily
available in a database (Figure 1A) (Jungbluth et al., 2007). However, if firstly the proteins are
separated by a wet milling process, and afterwards a wet starch stream is fermented to ethanol,
the study in the database would not be representative to analyse if this new setup is more
sustainable than the previous situation. The total impact of the end products would thus be
unknown, because the specific Unit Process Raw Data (UPR) with mass and energy data of this setup
is not available (Figure 1B), and because the parameters of the UPR in the database are fixed, and
cannot be adapted to the new situation.
1 A Basic Unit Operation (BUO) is a single engineering process such as a distillation, an evaporation
an oxidation, etc. It is NOT the combination of these Unit Operations to production chains which are
called Unit Operations in e.g. SimaPro
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Corn grains
Ethanol 95% in
H20 from corn at
distillery
Ethanol 99.7% in
H20 from corn at
distillery
DDGS from corn
Corn grains
Ethanol 95% in
H20 from corn at
distillery
Ethanol 99.7% in
H20 from corn at
distillery
DDGS from corn
Wet starch stream Protein slurry
Figure 1A (left) : ethanol from corn, available in a database. Figure 1B (right): ethanol from corn after protein separation, not available in a database (Jungbluth et al., 2007).
The assessment of new technologies therefore often results in so-called “data gaps” that should be
closed to allow a complete analysis of sustainability. Wernet et al. (2012) suggest that there are
four ways to obtain the required data:
Extrapolation from existing data
Substitution with generic/average datasets
Molecular structure-based models
Using process based estimation methods
Wernet et al. (2012) explain that the two first approaches should not be preferred because they
often result in an underestimation of the impact mainly because they are in many cases based on
the most simple products of that sector. Using molecular structure based models has its merits, but
is limited to specific sectors and follows a “black box” approach; as such giving little information on
the detailed impact of the production chain. The process based estimation methods are stated to
have the potential to deliver the best results for closing data gaps such as those occurring in
prospective sustainability assessments.
For this purpose, software such as ASPEN, SuperPro Designer and gPROMS can be used as they allow
changing parameters of the BUO and to simulate process chains used in new technologies. However,
they have several disadvantages for the application in LCA:
They are not open source
They are not compatible with the LCI databases: the output values are not always in the correct
units and are often exported to full reports instead of to a summary of Unit Process Raw Data
(UPR)
They do not offer the freedom of choice of complexity of process models. The calculations are
often very detailed, requiring many different inputs. Therefore they should be operated by
dedicated process engineers, who are often not familiar with LCA practices.
As such, the different sources of data mentioned are inadequate to obtain a full data inventory of
future technologies, hampering prospective assessments that are actually vital for sustainable
development. It is thus necessary to develop a process based approach that makes mass and energy
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data available before the construction of production plants in a format which can easily be coupled
to life cycle databases to simulate the background system. Such an approach would make it possible
to estimate sustainability at different system boundaries before large scale implementation of new
technologies. On the other hand, such an approach deals with the environmental and economic
dimension of sustainable development solely, as generic information on social impact is impossible
to obtain in advance because it depends more on company policy than on the technology itself.
Indeed, two exactly the same sets of unit operations necessary for a specific technology can have a
completely different social impact depending on the policy conditions.
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Goal
This basic engineering module approach aims at broadening the most frequently occurring type of
life cycle assessments, i.e. using existing studies or existing plant configurations added with fixed
data from databases, with a more generic approach which allows the modification of set, sequence
and type of BUO and process parameters in the foreground system that influence the life cycle mass
and energy inventory and thus the result of the sustainability assessment. This means that a set of
processes will be characterized in such a way that plants can be assembled together composed by
BUO, providing the attainable data necessary for prospective sustainability assessments, where the
specific focus of this work is on developing modules to estimate mass and energy flows of Basic Unit
Operations with a balance between complexity and accuracy. The modelling of the BUO of this
foreground system will make it possible to obtain inventory data of production chains of new
technologies compatible with life cycle assessment practices; i.e. easy to link to databases to model
the background system. This type of information, based on engineering approaches and rules of
thumbs, will be vital to know if new technologies will achieve a better performance than the
current ones and if they should be implemented or how they can be improved.
This work aims at a certain flexibility, by allowing the user to simulate the process chain with Basic
Unit Operations with different levels of work intensity (and correctness):
The production chain is fixed use existing studies from databases or using data from an
existing setup
The parameters of the production chain can be modified by the user
o The production chain can be modelled with engineering equations completed with
default values and input parameters from the user per BUO
o The production chain can be modelled with engineering equations completed with
detailed input parameters known by the user per BUO
Use simple rule of thumb values
With respect to Figure 1A, for example, the corn ethanol production chain can be taken from an
existing study from the database with fixed parameters. However, when in future the lignocelluloses
in the corn are co-fermented, the yield will change. This could be altered whilst working with
default parameters accounting for other factors such as electricity use if it is expected that the
setup is similar. Alternatively, if the user knows for example that a higher impeller speed, or a
longer residence time is necessary to obtain this co-fermentation this will influence the electricity
use and can therefore be modified. If changes are made, but very limited information is available,
simple rule of thumb values, if they are available, can be used to model the production.
This work is therefore subdivided in two parts. First, 25 different modules are developed and their
calculation procedures are explained. In the second part these modules are validated in a
biorefinery case study.
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Part I: Development of the engineering modules
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1. Principles
1.1 General
The structure of the Basic Engineering Modules is visualized in Figure 2. A distinction is made between
the foreground system α, which is the Basic Unit Operation under study, the foreground system β,
which includes the Supporting Unit Operations (SUO) necessary to support the foreground system α
within the factory gate (gate in to gate out supplied utilities/auxiliaries, e.g. cooling, pumping), and
the background system, γ, which accounts for the life cycle cradle to grave chain of these BUO. The
latter one is modelled by Life Cycle Inventory databases, whilst the foreground system α and β can be
modelled by the Basic Engineering Modules.
To model this complete system, the user will need to enter a minimal amount of input data:
Which BUO + characterisation
Which SUO + characterisation
Input flow characterisation at process gate in
Whilst the parameters for the latter are always similar, namely flow rate, composition, temperature
and pressure, the characterisation of the BUO and SUO depends on the type chosen.
If the minimal amount of input is entered, this can be completed with default values from the
databases DATABUO, which contains default characteristics of BUO, and DATAPHYSCHEM, which
contains default physical and chemical properties of resources and utilities. Alternatively, if the user
knows more detailed information of the setup, the used default values of these databases can be
edited. If too few information is available to use the calculation algorithms, simple Rule Of Thumb
values, from DATAROT database can be used.
Two types of output data are obtained:
Unit Process Raw Data; inventory data of the BUO in the α system and SUO in the β system, which
can be coupled to life cycle inventory databases to obtain a full LCI.
User output; mass and energy balances of the α and β system, which can serve as input to model
other BUO in the process chain of a final product or service.
As aforementioned, this approach is followed for the eco-efficiency assessment of future
technologies. This is visualized in Figure 3. Basic Unit operations need physical inflows (materials and
energy) to generate a certain amount of useful output and a certain amount of waste and emissions.
Similarly economic goods and services are needed to generate a certain amount of economic outflow.
On top of the costs of the resources, several other cost factors can be identified. This work focuses
on the horizontal (marked in red) structure and thus the mass and energy balances of the Unit
Operations, which can generate Unit Process Raw Data necessary for environmental and certain
aspects of the economic assessments.
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DATAPHYSCHEM:Database with editable
default physical and chemical properties of
utilities and resources such as chemicals and biomass
e.g. compositions, heat capacities, ...
Unit Process Raw data α (UPR) Mass and energy input and output of
the α system boundary
DATABUO:Database with editable
default characteristics of BUO
e.g. efficiencies, friction factors,...
Basic Engineering Modules: calculation algorithms
specific per BUO
Foreground system β:SUO: Supporting Unit Operations for
BUO α Typically cooling, pumping, ...
Unit Process Raw data β (UPR of gate in to gate out): Mass and energy input and output of the β system boundary
(Including α)
Life Cycle Inventory Database e.g. ecoinvent or
ILCD
Background system γ(Life cycle requirements necessary
for BUO α and β)
Life Cycle Inventory (γ)(cradle to gate out; including α and β)
DA
TAB
ASE
SD
ATA
INV
ENTO
RY
OF
SPEC
IFIC
CA
SEO
UTP
UT
β System boundary: gate in to gate out γ System boundary: cradle to gate out
Foreground system α:BUO: Basic Unit Operation
considered
α System boundary
User output
User input:- Choose BUO- Choose SUO
- BUO characterisation- SUO characteriztion
- Flow characterisation
Rule of thumb approach specific per BUO
DATAROT:Database with simple Rule Of Thumb values to model
BUO without need of additional information
Figure 2: the structure of the basic engineering modules, based on Van Der Vorst et al. (2009)
Basic Unit Operation with accompanying Supporting Unit Operations
(β system boundary)
Mass to be processed
(M)
Energy (En)
Auxilliary substances
(A)
Physical Inflow at gate in
Waste & Emissions
(WE)
Useful Output (UO)
Physical Outflow at
gate out
Economic Outflow (EO)
Labor (L)Equipment
(Eq)Other Costs
(OC)
Economic Inflow (EI)
Economic assessment
Environmental assessment
Figure 3: The environmental and economic aspects of the Basic Unit Operations
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For the purpose of the environmental assessment, the Physical Inflows, which can origin from the
technosphere or directly from nature, are categorized in 3 classes, with specific physical
characteristics:
Mass to be processed (M)
Mass flow
o Composition
o Temperature
o Pressure
Energy used (En)
o Electricity (for mechanical purposes)
o Other forms of energy:
Heating energy and/or heating medium
Cooling energy and/or cooling medium
Auxiliary substances (A)
Finally, these inputs will generate a mass and energy outflow, which consists out of useful/usable
products and energy (UO) and out of waste and emissions (WE) which can be emitted to nature, or
which can go back to technosphere. One unit of the useful output can be chosen as a functional unit.
Through the equations of the BUO the inflows, wastes and emissions can be linked to the functional
unit. By coupling this information to other BUO or LCI databases, data of the foreground and
background system can be obtained to obtain a full Life Cycle Inventory, which can then be assessed
for its impact.
1.2 Units
Preferred units are represented in Table 1.
Table 1: preferred quantitative and qualitative units used in the calculation tool
Unit\Category M Electricity Other forms of
energy A UO/WE
Quantitative kg kWh MJ kg kg, MJ or kWh
Qualitative (where relevant)
wt%
K
Pa
K
Pa
wt%
K
Pa
wt%
K
Pa
1.3 Time aspect
All Unit Operations are considered to be in continuous mode for the sake of simplicity and since
continuous operation is expected to be implemented more than batch mode in future industrial
development (f3factory.com, Reintjes, 2009). The unit of time chosen is seconds, thus all units
presented in Table 1 are expressed per second (MJ becomes MW). The different flows can thus be
presented per time or per functional unit.
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2. Unit operations included
The BUO are subdivided in 4 major classes based on their functionality. The calculations will focus on
the industrial application of these BUO.
Reactions
o Chemical reactions
o Incineration for heat and power
o Fermentation
Separation processes
o Evaporation
o Distillation
o Filtration
o Sedimenting centrifuges
o Electrostatic precipitation
o Electrodialysis
o Pressure swing adsorption
Physical mechanical processes
o Mechanical compression – single stage
o Mechanical compression – multi stage
o Pumping incompressible fluids
o Pumping incompressible fluids through packings
o Agitation and mixing of liquids and suspensions
o Comminuting
o Fluidization, pneumatic drying and pneumatic conveying
o Conveying solids
o Fans, blowers & vacuum pumps
Utilities
o Heating
o Cooling
o Steam generation
Processing and use of materials
o Electric manufacturing processes
o Material requirements as a function of mechanical properties
o Fuel use of cars
Each BUO focuses on the process within the α system boundary and generates UPR data for this
system. Supporting Unit Operations of the β system should therefore be considered separately, whilst
flows in the γ system boundary should be taken from a life cycle inventory database.
2.1 Reactions
2.1.1 Chemical reactions
Chemical reactions, are modeled with the thermodynamic equilibrium model. This is a very useful
approach for generic modelling, since it is stated that this approach is independent of reactor design
(Puig-Arnavat et al., 2010). However, for each specific case, several assumptions will have to be
made. Most importantly: a thermodynamic equilibrium should be reached, which is not always the
case. Therefore, residence time in the reactor should be high enough. On top of this, the process is
assumed to be adiabatic (no heat losses), the conditions in the reactor should be constant without
spatial variation and gaseous products are assumed to be ideal. The model does not include effects
with byproducts such as tar, ash, micro-organisms, etc. However, for the latter the amount of such
byproducts formed can be deducted from the original amount of reactants if this quantity is known.
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System boundary
The α system boundary for this BUO includes the reaction of the reactants to products in gas or liquid
(including slurries) state in any type of vessel that allows a thermodynamic equilibrium. It is valid
over a broad temperature range, as long as the coefficients for the heat capacity calculations remain
valid; often between 200 and 1500K. Apart from the quantity of output products (useful, UO, and
wastes, WE), the module calculates the amount of heating or cooling energy, if required (in MJ).
Possible SUO are mixing, pumping and cooling. Heating can be found in LCI databases, or can be
modelled by the BUO ‘heating’, ‘steam generation’ or ‘incineration for heat and power’.
Chemical reaction
α System boundary
Reactants (M)
Products (UO)
β System boundary
Mixing Pumping Cooling
Heating (En)
Products (WE)
Calculation algorithm
Three type of equations are used to solve chemical reactions of which the reactants and reaction
products are known. Firstly an elementary balance can be performed. If this is not sufficient, extra
equations can be added by considering the thermodynamic equilibrium of the different reaction
mechanisms. The third type of equation is solving the enthalpy balance of the reaction, which can be
used to obtain additional stoichiometric information or to have an estimate of temperature.
1. Elementary balance
2. Equilibrium constants (Coker, 2001): temperature dependent
The temperature dependent Gibbs free energy of reaction can be obtained with the enthalpies and
entropies of the different products and reactants:
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With ΔH the temperature dependent heat of formation, and ΔS the temperature dependent molar
entropy.
Where the heat of formation can be obtained by:
The standard heat of reaction ( ) can be:
Input from user
Calculated
With αp and αr the stoichiometric coefficients.
Without phase changes the enthalpy difference between the products and reactants can be based on
the heat capacity:
With T the actual temperature and T0 the reference temperature (298K).
With a, b, c and d the regression coefficients of the heat capacity temperature dependence equation,
which can be found for many chemicals in ‘The Chemical Properties Handbook’ (Yaws, 1998). Thus:
This value represents the enthalpy change of the reaction, and can therefore be used to calculate the
amount of heating or cooling (q) that is required to keep a constant temperature of reaction.
Similarly, the entropy balance can be calculated by:
With:
Finally, the enthalpy balance of the reaction can be formulated:
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User and database input
Input values required from user DATAPHYSCHEM DATABUO
Mass flow rate and type of feed
Reaction products
Estimated temperature
Elementary composition feed
Standard enthalpy and entropy of
products and reactants
Heat capacity coefficients
-
User output
Output product flows
Temperature of output flows
Cooling energy required (in MJ, to be calculated by cooling module)
Heat required
UPR output
From technosphere:
Heat required if necessary (in MJ), e.g. “heat, natural gas, at boiler atmospheric non-modulating
<100kW”
Quantity of inputs necessary for certain products, e.g. 1 MJ syngas requires 0.074kg wood chips.
Example
A gasification produces an energy gas (syngas) consisting mainly out of hydrogen, carbon monoxide,
carbon dioxide, water vapor, methane and nitrogen.
Where CHxOyNz is a type of reactant (M) (fuel, e.g. coal or biomass), w is the moisture content of the
fuel, m is the used amount of air per mole fuel and x1 to x5 represent the molar composition of the
syngas.
Different reactions occur in the gasification zone, which can be combined to two :
Boudouard reaction: (A)
Water-gas reaction (B)
Methane reaction (C)
(A) and (B) can be combined to the Water gas-shift reaction:
(1)
And (B) and (C) to the methane synthesis reaction:
(2)
In the following, thermodynamic equilibrium is assumed in the gasification zone. This is often done,
however, it might not always be the case at lower temperatures (750-1000°C) (Puig-Arnavat et al.,
2010).
The equilibrium constants for the resulting equations are:
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The 5 equations with 5 unknowns are thus the carbon balance, the hydrogen balance, the oxygen
balance, and the two equilibrium reactions.
Additionally, air necessary (m), or temperature T can be calculated by:
With Hf the enthalpies of formation of the respective components.
As a rule of thumb form, the relationship of Borman and Ragland for combustion can be used (Borman
& Ragland, 1998):
If the fuel has a composition of CaHbOcNd then the air fuel ratio is equal to a + 0.25b-0.5c.
Where a gasification is a partial oxidation, needing 30-60% of this stoichiometric air amount (Vaezi et
al., 2008).
For example if 1000K and an air/fuel ratio of 0.4 is assumed, it can be calculated how much fuel is
needed to obtain the functional unit of a syngas with a LHV of 1MJ.
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Figure 4: Example of a gasification at 1000K with an air/fuel ratio of 0.4 and biomass fuel with 90% dry matter (DM) content
The UPR of 1MJ syngas from wood chips, where it is assumed that no additional heat is added, would
thus be:
From technosphere:
0.074kg wood chips (90%DM)
2.1.2 Incineration for heat and power
Incineration is a technology which is typically used to gain energy from different types of resources by
oxidation. The engineering module calculates the energy in the form of heat, electricity or a
combination, generated in different types of industrial combustion configurations (Table 2). In the
first step, the Lower Heating Value (LHV) of the energy source(s) is calculated, whilst in the second
step efficiencies to convert LHV to electricity (and steam) are accounted for. Other calculated
outputs are: the auxiliary power needed, ash production and emission and a rough estimation of the
other main flue gas components H2O, CO2, NOx, SO2, P2O5.
System boundary
The BUO “incineration for heat and power” includes the reaction of the incoming fuel in an industrial
combustion chamber, and the optional conversion of the flue gas to steam and electricity in a boiler
and/or turbine and generator. A certain amount of power is needed to support the combustion
process in the β system boundary. Literature numbers found vary from 2%(Henderson, 2004) to 6%
(Bedi) of total feed LHV. A value of 3% is chosen (Kehlhofer et al., 2009) which is subtracted from LHV
before calculating the energy output.
21
Combustion chamber
α System boundary
Fuel (M)
Electricity (UO)
β System boundary
Air compressor/
blower
Pumping/conveying
fuel
BoilerTurbine
GeneratorSteam (UO)
Hot flue gas (UO or WE)
Pumping water
Table 2: Included setups for energy generation
Steam boiler
Small burner for electricity (all inputs)
Large burner for electricity (all inputs)
Efficient burner for electricity with combined cycles (fossils)
High Efficient burner for electricity with combined cycles (gas)
CHP (all inputs)
Efficient CHP (fossils)
High efficient CHP (gas)
Calculation algorithm
For fossil fuels standard LHV’s are available in a database, whilst for organic chemicals (e.g. waste
solvents) standard net enthalpies of combustion are documented in literature. These values can be
found in DATAPHYSCHEM.
For coal and biomass standard values can be chosen from DATAPHYSCHEM as well, but since their
composition can differ substantially the LHV can be calculated in MJ/kg (dry weight) based on the
ultimate analysis of the fuels.
For coal this is calculated with so-called Milne formula, which was specifically developed for coal and
tested on real data from different coal types (Institute of Gas Technology, 1978; Hoinkis & Lindner,
2007):
For biomass this can be calculated with the most appropriate equation for biomass (R² = 0.834)
(Sheng & Azevedo, 2005):
Where the elemental compositions are in wt% and where:
22
The database contains default elementary compositions (based on dry weight) for different types of
coal and biomass.
The LHV is in both cases calculated by (Phyllis; Fowler et al., 2009):
The LHVwet can be correlated to the LHVdry by:
With MC the Moisture Content of the wet feed in wt%. The LHVwet value in MJ/kg is then multiplied
with the feed flow in kg/s to give the total LHVwet in MW.
If the energy source is a mixture of fuels, it is assumed that the LHV can be added linearly for input
mixtures. For example for two input streams x and y:
Based on elementary composition, either from the database or from the user, a rough calculation can
be made to estimate the main flue gas components H2O, CO2, NOx, SO2, P2O5. A complete oxidation of
the basic components is assumed for simplicity and the emissions are expressed in kg/s.
Thus making a mass balance:
1kg H 9kg H2O
1kg C 3.67kg CO2
1kg N 3.29kg NOx
1kg S 2kg SO2
1kg P 2.29kg P2O5
CO formation depends on local temperature differences in the combustion chamber and therefore it
is very difficult and data intensive to make generic models for CO formation (Velicu & Koncsag, 2009).
For the mass balance, NOx is calculated as NO2. The main NOx component formed at the incineration
is NO, but this is instable and will react further with oxygen to NO2 when released in air. Only fuel NOx
is considered: the Zeldovich mechanism accounting for thermal NO formation from N2 originating
from fuel or air is not considered, because of uncertainties in the currently existing models and high
required input data (the reactions depend amongst others on residence time, local oxygen
concentrations and local temperature) (Schwerdt, 2006).
Ash production is calculated for biomass and coal based on data from the ultimate analysis in the
database (DATAPHYSCHEM). Of this ash content, 98% is retained, whilst 2% is emitted in the case an
ash filter installation (Röder et al., 2004). If this is not the case, 80% is retained, and 20% is emitted
(Schobert, 2002).
To predict the amount of air that is used, the relationship of Borman and Ragland can be used
(Borman & Ragland, 1998):
If the fuel has a composition of CaHbOcNd then the air fuel ratio is equal to a + 0.25b-0.5c.
User and database input
Input values required from user DATABUO DATAPHYSCHEM
Mass flow rate and composition of feed
Type of energy generation setup
Ash filter; yes/no
Efficiencies
Ash retention
Enthalpies of combustion
Ultimate analysis of the fuel
23
User output
Energy output in MJ as
o Flue gas and/or
o Steam or/and
o Electricity
Ash retention
Air needed
UPR output
From technosphere:
Fuel needed for 1MJ energy
Disposal:
Ash
To nature:
Emissions of ash, H2O, CO2, NOx, SO2, P2O5
Example
To obtain 1 kWh electricity in a small CHP, following UPR is obtained:
From technosphere:
0.828kg “wood chips, hardwood, from industry, u=40%” is necessary.
Disposal:
0.01kg wood ash mixture
Byproducts (not allocated yet)
6.54MJ steam
To nature
0.00021kg PM
0.449 kg H2O
1.44kg CO2
0.013kg NOx
This UPR can thus be allocated to the electricity and steam. The UPR of the SUO (β-α) can be
estimated as 0.12 kWh electricity.
2.1.3 Fermentation
The fermentation includes the conversion of a starch/glucose feed stream to a product/water/rest
products/biomass mixture and optionally a CO2 gas stream. Calculations are made by using extended
Haldane kinetics, which includes substrate affinity, substrate inhibition and product inhibition.
However, it does not include biomass decay and maintenance.
System boundary
The α system boundary includes only the conversion of the feed stream in the fermentor. The module
also calculates the mixing time and the potential need of cooling or heating, which can be used to
calculate the UPR of the SUO in the β system boundary. Most relevant SUO are thus mixing, pumping
and heating/cooling.
24
Fermentation
α System boundary
Feed (M)
Products (UO)
β System boundary
Mixing Pumping Cooling
Heating (En)
CO2 emissions (WE)
Nutrients (A)
Calculation algorithm
If the feed stream is starch, it is assumed to be completely hydrolyzed according to:
This means that the mass of starch should be multiplied with 180/162 to include the additional water
in the final mass of glucose.
The glucose available is then fermented to an end product according to one of the reactions in Table
3, depending on the reaction conditions and type of micro-organism used:
25
Table 3: Stoichiometric fermentation reactions and current and future product yields per substrate consumed (Patel et al., 2006)
Product Reaction products
Product yield; YP/S
(g product/g substrate)
C F
Ethanol 0.46 0.47
PDO (1,3 – Propanediol) 0.41 0.54
ABE (butanol) 0.42 0.50
Acetic acid 0.50 0.90
Acrylic acid 0.72
Lactic acid 0.93 0.95
Succinic acid 0.88 1.01
Adipic acid 0.17 0.47
Citric acid 0.86 0.96
*C = Current, F = Future
26
The generic reaction which occurs when 1 mole of substrate is consumed includes the formation of
biomass
Where R is the amount of moles carbon in the reaction specific rest products, which can be found in
the reaction equations of Table 3, and B is the amount of moles of biomass which is formed per mole
of substrate consumption and which is based on the carbon balance:
With YP/S is the yield of product per substrate (g/g), which can be found in Table 3. The rest products
are reaction specific (see Table 3). The elementary balance of hydrogen and oxygen will not be
checked due to the complexity of bacterial growth. Therefore standard values of bacterial
composition are taken, which can be found in Table 4 (Harding, 2008). Based on these compositions,
the amount of nitrogen, sulphur and phosphorus nutrients to be added in the fermentor can be
calculated.
Table 4: Elemental formula for micro-organisms (Harding, 2008)
Organism C H O N S P
Aerobacter aerogenes 1 1,83 0,55 0,25
Aspergillus niger 1 1,74 0,711 0,117
Azohydromonas lata 1 1,76 0,48 0,19
Candida sp. 1 1,84 0,52 0,16
Escherichia coli 1 1,77 0,49 0,24
Klebsiella sp. 1 1,75 0,43 0,23
Paracoccus denitrificans 1 1,66 0,49 0,2
Pseudomonas C12B 1 2 0,52 0,23
Saccharomyces cerevisiae 1 1,76 0,53 0,17 0,005 0,01
Other 1 1,82 0,53 0,2
The total output mass flow (kg/s) of end product formed is:
With Sin the mass flowrate of substrate in (kg/s) and Subeff the fraction of input substrate converted:
Where Sused is the substrate fermented and Subeff is input from the user, which is taken as a default
value of 90%. However, the user can vary this efficiency to follow the effect on the reaction time
The unused substrate (Sun) is determined by:
27
In continuous mode, the enthalpy change of the reaction allows the calculation of amount of cooling
or heating medium required to maintain a certain temperature. For example for ethanol fermentation
by Saccharmoyces cerevisiae, following reaction enthalpy can be used:
(Chongvatana, 2007-2008)
To determine the reaction time, and thus reactor volume, necessary for the fermentation, Haldane
kinetics with product inhibition can be used. The rate of biomass formation (rb) can be calculated by
(Ghose & Tyagi, 1979):
Where the specific growth rate, µ (h-1), is determined by:
B = biomass concentration (g/l), Ks Monod constant (g/L), KI inhibition constant (g/L), µmax the
maximum specific growth rate (h-1), P the ethanol concentration in the fermentor (g/l), α the product
inhibition constant, Pmax the maximum ethanol concentration (g/l) and CSun = substrate concentration
(g/L). A perfectly mixed fermentor is assumed, implying that the concentration in the tank is the
same as the concentration in the outlet flow. In a continuous reactor, this can be determined by:
With M and ρ the mass and density (kg/l) of the unused product (un), the water (H2O), the end
product (prod), the biomass (B) and the rest products respectively and assuming that density can be
added linearly.
These kinetics can be simplified to
in the case that substrate and product inhibition are negligible, and if substrate concentration is high
in comparison to the Monod constant. However in industrial practice this might not always be the
case, since it is the goal to consume as much as the substrate as possible to form high concentrations
of end products. Furthermore, the product inhibition factor and substrate inhibition
factor can be neglected if no influence of product and substrate concentration is expected.
Based on these reaction rates, the residence time in the reactor can be estimated:
With MMS the molar mass of the substrate and V the volume of the tank (in l):
Reaction time can thus be determined by:
Where Q is the flow rate in (l/s):
28
User and database input
Input values required from user DATABUO DATAPHYSCHEM
Fermentation reaction
Type of micro-organism
Mass and composition of input flow
Substrate conversion efficiency
Yield coefficient
Kinetic parameters
Reaction enthalpy
(preferably entered by user)7
Composition feed streams
User output
End product/water/rest product/biomass sludge composition
Reaction time
Cooling/heating energy required (MJ)
UPR output
From technosphere:
Amount of biomass needed for 1 kg fermentation product
Kg nitrogen, sulphur and phosphorus containing product required per kg fermentation product
To nature:
Kg biogenic CO2 emitted
Example
A glucose stream with 1kg/s glucose and 10kg/s water enters a fermentation section with
Saccharomyces cerevisiae to form ethanol. When a substrate consumption efficiency of 90% is taken,
and with a default yield coefficient of 0.46kg/kg, 0.414kg/s ethanol and 0.396kg/s CO2 is formed,
whilst 0.1kg/s substrate remains unused.
Using the elementary carbon balance, 0.6 moles biomass (CH1.76O0.53N0.17S0.005P0.01) is formed per mole
glucose fermented. This means a total mass flow of biomass of 0.075kg/s. As such, 0.102 mole N,
0.003 mole S and 0.006 mole S should be added per mole glucose fermented or 0.51, 0.015 and 0.03
moles nutrients in total. Therefore, 15.3g ureum, 1.47g H2SO4 and 4.26g P2O5 is needed.
The substrate concentration in the fermentor can be determined based on this unused substrate and
the output flow. Csun thus becomes 10.65g/l.
With default chosen kinetic parameters for ethanol production (µmax = 0.584, KS = 0.155g/l, Ki
=160.7g/l, Pmax = 125g/l, α = 3.68), µ becomes 0.138hr-1 and with a biomass concentration of 10g/l, rp
becomes 1.38g/l/h.
Based on the total mass of biomass formed, and on this biomass formation rate, a residence time of
5.11hr is calculated.
Rescaling to the functional unit of 1kg ethanol, the UPR α becomes:
From technosphere
2.41kg glucose, or if glucose is estimated as 60wt% of corn, 4kg corn
37.0g ureum
3.6gH2SO4
10.3g P2O5.
To nature
0.96kg carbon dioxide, biogenic
29
2.2 Separation processes
2.2.1 Evaporation
The module ‘evaporation’ calculates the energy required to concentrate a solution consisting of a
non-volatile solute and a volatile solvent. Where the latter is water in the majority of the cases. In
contrary to drying, the residue of an evaporation is a (sometimes highly viscous) liquid, whilst the
main difference with distillation is the fact that no efforts are made to obtain a concentrated vapour
(McCabe et al., 2004). It can be used for 1 to 3 effect evaporators and for evaporators with
mechanical recompression.
System boundary
The BUO ‘evaporation’ includes the 1 to 3 effect evaporator. Pumping operations for the different
liquid and gas streams, and vacuum pumps are not included in the α system boundary. Furthermore,
recompression should also be added as a SUO if this is necessary. Calculations are valid over a broad
temperature range, as long as the coefficients for the heat capacity calculations remain valid.
Evaporation
α System boundary
Thin liquor (M)
Thick liquor (UO)
β System boundary
Pumping
Heating (En)
Vapor (UO/WE)
Recompression
Condensate (UO/WE)
Vacuum pump
Calculation algorithm
The heat required for the evaporation can be obtained by making an enthalpy balance, with a thin
liquor coming in (f) and thick liquor going out (o):
With m the respective masses, Hv the specific enthalpy of the evaporated vapor, Hf the enthalpy of
the incoming feed, Ho the enthalpy of the thick liquor and η an efficiency factor, typically 0.9,
accounting for heat losses.
30
Or shorter
With λ the latent heat of vaporization of the evaporated vapor, cp the mean heat capacity of the
feed, Tb the boiling temperature of the mixture and Tf the temperature of the feed.
This model assumes a negligible heat of dilution, which is typical for flows containing sugars, organic
salts and papermill liquors (McCabe et al., 2004).
In order to save energy, a part of the enthalpy in the evaporated vapour is recovered in sequencing
vessels or ‘effects’, which is called multi effect evaporation. To estimate the amount of heat savings
per effect, the default values in Table 5 from DATABUO can be used for one to triple effect
evaporators (Grosse & Duffield, 1954). The heat required then becomes:
Table 5: Default steam savings in one to triple effect evaporators
Amount of effects Calculated amount of heat required per amount evaporated (Feff)
1 effect 1.00
2 effects 0.52
3effects 0.37
User and database input
Input values required DATABUO DATAPHYSCHEM
Mass and type of feed
Feed temperature
Pressure
Mass of water to be evaporated
Number of effects
Recompression; yes/no
Steam savings for 1 to 3 effect
evaporators
Heat capacities
Enthalpy of vaporization
Boiling temperature
User output
Output product flows
Temperature of output flows
Need for recompression
Heat required
UPR output
From technosphere:
Heat required (in MJ)
Example
A wet starch stream of 29000L water enters at 20°C and is evaporated in a single effect evaporator at
1atm to a solution 9000L. Thus 20000L of water is evaporated. In a regular evaporation unit with Hv =
2257kJ/kg, cp, water = 4.17kJ/kgK, cp, starch = 1.53kJ/kgK and Tb = 100°C, this results in a heat
requirement of 54993MJ.
The UPR of evaporating 1L water from this stream is thus:
From technosphere:
2.75MJ heat, unspecific, in chemical plant
31
Rule of thumb values
In the BREW study (Patel et al., 2006) literature values for single-stage- and multi-stage evaporation
have been collected. Steam requirements for single-stage evaporation range from 0.005-1.4 kg
steam/kg evaporated. The recommended value was determined at 1.2 kg steam/ kg evaporated.
Electricity use for single-stage evaporation was estimated at 0.04 kWh/ kg evaporated. For multistage
evaporation, literature values ranged from 0.01-1.25 kg steam / kg evaporated. The recommended
value was determined at 0.1-0.5 depending on the amount of stages. Electricity use for multistage
evaporation was estimated at 0.002-0.0344 kWh/ kg evaporated. The recommended value was
determined at 0.005 kWh/kg evaporated.
2.2.2 Binary distillation
Distillation is used as a separation process based on different boiling points. In contrary to
evaporation, all components are appreciably volatile. Furthermore, the vapour phase is condensed in
an overhead condenser, optionally with a reflux, the latter returning a part of the condensed vapour
to the distillation column.
System boundary
The BUO can be used for non extractive binary distillations with or without reflux. It calculates the
heat required for the evaporation process (often in a reboiler) and the cooling energy required in the
condensation (incl. reflux). This is valid over a broad temperature range, as long as the coefficients
for the heat capacity calculations remain valid. Pumping operations for the different liquid and gas
streams, vacuum pumps and possible mixing operations are not included in the α system boundary.
Furthermore, cooling (in the condenser) should be added as a SUO in the β system boundary.
Distillation column
α System boundary
Distillation feed (M)
Bottoms (UO/WE)
β System boundary
Pumping
Heating (En)
Distillate (UO/WE)Overhead condensor
CoolingVacuum
pump
V
L
Mixing
Calculation algorithm
The general equation for calculating the energy requirements of a distillation is very similar to the
equation for an evaporation, but includes the fraction of reflux that has to be added to the enthalpy
of vaporization:
With mf and mb the mass of the feed and the bottoms respectively, λv the latent heat of the vapor, cp
the mean heat capacity of the feed, Tb the pressure dependent boiling point of the vaporized
component and η an efficiency factor, typically 0.9, accounting for heat losses.
32
The Reflux Ratio (RR, or L/V) can be calculated based on the minimum reflux ratio (RRmin) with a rule
of thumb (Perry & Green, 1999):
Assuming negligible holdup of liquid on the trays, in the column, and in the condenser, the minimum
reflux ratio can be obtained by:
With ypi and xpi the initial molar fraction of the more volatile component and yd and xd the molar
fraction of the more volatile component in the distillate in the vapour phase and liquid phase
respectively. Where xd and yd are equal assuming a total condensation (Figure 5 (Perry & Green,
1999)).
Figure 5: Determination of the minimum reflux ratio from the Equilibrium curve and operating line in a distillation. Point D is the composition of the distillate, whilst point F is the composition of the feed (Perry & Green, 1999)
Assuming that the vapor is cooled to obtain a phase change, the cooling duty for the condensation
(qc) of is:
Aforementioned theory is generally applicable, however does not include the specific cases of
azeotropes. A frequently used example is the case of an ethanol – water – biomass mixture obtained
after a fermentation. To include this case it has been modelled using SuperPro Designer based on the
amount of wt% ethanol in the fermentation slurry.
The equation obtained is:
33
With y the amount of MJ steam required per kg ethanol and x the weight percentage bioethanol in
the slurry.
Figure 6: the heat required per kg ethanol for the ethanol distillation (at 85°C) in MJ/kg ethanol after a fermentation to obtain 94% ethanol/water
User and database input
Input values required DATABUO DATAPHYSCHEM
Composition feed
Compositions distillate -
Heat capacities
Enthalpy of vaporization
Pressure dependent boiling
temperature
User output
Output product flows
Temperature of output flows
Cooling energy required (MJ)
Heat required
UPR output
From technosphere:
Heat required (in MJ)
Example
A mixture of 800 kg in 9000kg water, 100kg biomass and a substrate leftover of 100kg glucose means
8% ethanol weight percentage resulting in a heat requirement of 4.63MJ/kg. Assuming a reflux ratio
of 0.9 (RR=L/V), the cooling requirement becomes 1.60MJ/kg.
34
The UPR α of distilling 1kg ethanol to 94% is thus.
From technosphere:
4.63MJ heat, unspecific, in chemical plant
Rule of thumb values
In the BREW study (Patel et al., 2006) literature values for distillation have been collected ranging
from 0.9-4.4 kg steam/kg of product and 0.07-0.14 MJe/kg of product. No recommended value was
suggested but instead the energy use for distillation should be determined by multiplying the heat of
evaporation by a reflux factor of 1.3. Electricity demand to generate vacuum, which may reduce the
heat requirements of the distillation column are assumed to be very low to negligible.
2.2.3 Filtration
In chemical processes, filtration is the mechanical or physical operation which is used for the
separation of solids from fluids (liquids or gases) by leading them through a medium that allows only
the fluid to pass and that retains oversize solids. As a filtration medium, normally a solid sieve or a
membrane (surface filter) is applied although filtration can also occur through a bed of granular
material (depth filter). Fluids flow through a filter due to a difference in pressure – fluids flow from
the high pressure side to the low pressure side of the filter, leaving solids behind. The application of
gravity is the simplest way to achieve this, but in the laboratory, pressure in the form of compressed
air on the feed side or vacuum on the filtrate side may be applied to enhance the filtration process.
In industry, when a reduced filtration time is important, the liquid may flow through a filter by the
force exerted by a pump.
System boundary
The α system boundary for this BUO includes the filtration of a suspension and the pumping operation
applied to create a pressure difference. Depending on the requirements, either the filtrate or the
solids are the useful output.
User and database input
Calculation algorithm
The power for filtration (Pf in W) is calculated according to (Harding, 2008):
Filtration
System boundary
Pumping
System boundary
Suspension
(M)
Filtrate
(UO/WE)
Solids
(UO/WE)
Electricity
(En)
35
In which:
P = Pressure difference across the filter (Pa)
Q = Flow through the filter (m³/s)
eff. = Pumping efficiency (dimensionless)
User and database input
Input values required DATABUO DATAPHYSCHEM
Flow of fluid Pressure difference Pumping efficiency
User output
-
UPR output
From technosphere:
Amount of electricity required, e.g. “Electricity, medium voltage, production UCTE, at grid”.
Example
An industrial filtration unit is assumed with a filtration flow 1 m3/s. The pressure difference is 1000
kPa and the pumping efficiency is 0.8 (Harding, 2008). In this situation, the required power for
filtration is 1250kW, or 0.35kWh/m³.
Rule of thumb values
In the BREW study (Patel et al., 2006), Values for different types of membrane filtration have been
estimated. An overview is given below:
Type of membrane filtration Unit Value range Chosen value
Microfiltration kWh/m3 permeate 1.2-2.6 2
Ultrafiltration kWh/m3 permeate 3.5-16 5
Diafiltration kWh/m3 permeate 5 5
Nanofiltration kWh/m3 permeate 1-7 7
Reverse osmosis kWh/m3 permeate 2.5-10 9
2.2.4 Sedimenting Centrifuges
Sedimenting centrifugation is a broadly used, but energy intensive equipment to separate solids from
liquids based on a difference in density, without using a filter (Perry & Green, 1999).
System boundary
This BUO focuses in the α system boundary on the electricity use of the centrifuges. Data is available
for tubular, disk, disk with nozzle discharge and decanters/helical conveyors. Necessary pumping
operations are situated in the β system boundary. An approximation can be made of the separation
efficiency, to serve as input for further BUO.
36
Sedimenting centrifuge
α System boundary
Feed (M)
Solids (UO)
β System boundary
Pumping
Electricity (En)
Centrate (UO/WE)
Calculation algorithm
Due to the complexity of the process, where the settling speeds and power consumption depends
heavily on type of centrifuge and particle size and shape of the solids, it is very difficult to obtain
one generic equation. Therefore, the power consumption is modeled by using default powers of
different types of equipment, which can be taken from DATABUO (Table 6 (Perry & Green, 1999)).
Separation efficiencies can be calculated with the sigma (Σ) concept, starting from the solids
concentration in the underflow stream, which can be estimated by (Van Der Meeren, 2004; Brennan,
2006; Woon-Fong Leung, 2007):
Table 6: Typical solids concentrations in the underflow of a centrifugation
Type Typical solids concentration (volume%)
Tubular 95%
Disk 40%
Disk; nozzle discharge 40%
Decanter / Helical conveyor 95%
Whilst the efficiency of the separation can be based on the maximal theoretical flow (m³/s) rate at
which a 100% separation is reached (Axelsson & Madsen, 2006):
With ε an efficiency factor, correcting the theory for practical application, where this factor is 95%
for tubular bowl centrifuges, 60% for decanter centrifuges, and 60% for disk centrifuges.
v is the settling velocity (m/s):
Dp is the diameter of the solid particle (m) to be separated, µ is the dynamic viscosity of the liquid
(Pa.s), g the gravity constant (9.81m/s²) and ρp and ρ are the densities (kg/m³) of the particle and
the liquid respectively.
37
Σ is the sigma factor, depending on the type of centrifuge used:
For disk-bowls
Where r2 is the outer radius of the disk, and r1 the inner radius of disk, N is the number of disks and θ
is the half cone angle of the disks. w is the rotational speed (rad/s):
With n in rpm, which can be taken from Table 7.
For tubular centrifuges:
With R the radius of the bowl (m), and ri the radius of the inner layer of the fluid (m), thus R-ri being
the thickness of the fluid in the centrifuge.
For decanters:
With L1 the length (m) of the cylindrical part, and L2 (m) the length of the conical part.
38
Table 7: Typical centrifuge configurations and power (pw) (Perry & Green, 1999)
Type Bowl diameter
(cm)
Typical speed
(r/min)
Maximal centrifugal
force x gravity
Liquid throughput
min (l/s)
liquid throughput
max (l/s)
Solids
throughput min
(kg/s)
Solids throughput
max (kg/s)
Typical
motor size
(kW)
Tubular
4.45 50000 62400 0.003 0.016
2
10.48 15000 13200 0.006 0.630
1.5
12.70 15000 15900 0.013 1.260
2.2
Disk
17.78 12000 14300 0.006 0.630
0.2
33.02 7500 10400 0.315 3.150
4.5
60.96 4000 5500 1.260 12.600
5.6
Disk; nozzle
discharge
25.40 10000 14200 0.630 2.520 0.028 0.278 14.9
40.64 6250 8900 1.575 9.450 0.111 1.111 29.8
68.58 4200 6750 2.520 25.200 0.278 3.056 93.3
76.20 3300 4600 2.520 25.200 0.278 3.056 93.3
Decanter / Helical
conveyor
15.24 8000 5500
1.260 0.008 0.069 3.7
35.56 4000 3180
4.725 0.139 0.417 14.9
45.72 3500 3130
6.300 0.278 0.833 37.3
60.96 3000 3070
15.750 0.694 3.333 93.3
76.20 2700 3105
22.050 0.833 4.167 149.2
91.44 2250 2590
37.800 2.778 6.944 223.8
111.76 1600 1600
44.100 2.778 6.944 298.4
137.16 1000 770
47.250 5.556 16.667 186.5
2 Turbine driven, 372kPa necessary (steam or air compressor)
39
User and database input
Input values required DATABUO DATAPHYSCHEM
Type of centrifugation
equipment
Liquid/solid throughput
Centrifuge size
characterisation
Centrifuge power
Rotational speed
Bowl diameter
Viscosities
Densities
User output
Output product flows
UPR output
From technosphere:
Electricity used (kWh)
Example
A stream of 20kg/s water and 0.5kg/s solids (assuming density of solids 1300kg/m³ and the
viscosity of the fluid 0.005Pa.s) is centrifuged in a disk centrifuge with nozzle discharge. This
requires a centrifuge of 93.3kW.
If the assumed rotational speed is 4200rpm, in a disk with nozzle discharge centrifuge of 76cm
diameter, with 30 disks at 35° the sigma factor becomes approximately 30000m², and the
maximum flow to obtain a centrate stream free of solids of 0.1mm diameter is 5.76m³/s. This
means that in this situation a pure water stream of 19.42kg/s is obtained and a underflow
containing 0.5kg solids and 0.58 kg water per second.
The UPR α of separating 1kg solids from 97.5 to 60%v/v moisture content is thus:
From technosphere
0.052kWh electricity, medium voltage
Rule of thumb values
In the BREW study (Patel et al., 2006) literature values for centrifugation have been collected for
yeast centrifugation and for bacteria centrifugation. The values for yeast centrifugation range
from 0.7-2.5 kWh/m3 of feed. The recommended value was determined at 1.5 kWh/m3 of feed.
The values for bacteria centrifugation range from 6.2-25 kWh/m3 of feed. The recommended
value was determined at 7 kWh/m3 of feed.
2.2.5 Electrostatic precipitation (ESP)
Electrostatic precipitators use an induced electric charge to separate solids from a gaseous
stream.
System boundary
This BUO includes the electricity used for the corona generation and is valid for gas velocities
between 0.5 and 3m/s. The module is aimed at dry electrostatic precipitation of particulate
matter from 0.01 to 10µm. It does not include pumping operations.
40
Electrostatic precipitation
α System boundary
Gas (M)
Cleaned gas (UO)
β System boundary
Pumping
Electricity (En)
Removed PM (WE)
Calculation algorithm
The electrical corona produced requires a certain amount of power, which can be calculated by
(University of Florida, Environmental Engineering Sciences, Aerosol & Particulate Research Lab,
2011):
P is the Power (in W), Q is the volumetric gas flow (in m³/s) and η is the efficiency of the
precipitation in %.
User and database input
Input values required DATABUO DATAPHYSCHEM
Gas flow rate + composition
Efficiency of solids removal - -
User output
Mass of solids precipitated
UPR ourput
From technosphere:
Electricity use
Example
A polluted stream of 2m³/s contains 500g/m³ particulate matter. To achieve a removal efficiency
of 95%, 0.116kWh/h electricity is consumed. 950g/s of pollutant is precipitated.
The UPR α of cleaning 1m³ gas is thus:
From technosphere
1.61E-5 kWh electricity, medium voltage
2.2.6 Electrodialysis
Electrodialysis is a separation technique that uses an applied electric potential difference for the
transportation of salt ions from one solution to another through ion-exchange membranes. This is
done in a configuration called an electrodialysis cell. The cell consists of a feed (dilute)
compartment and a concentrate (brine) compartment formed by an anion exchange membrane
41
and a cation exchange membrane placed between two electrodes. Under the influence of an
electrical potential difference, negatively charged ions in the dilute stream migrate towards the
positively charged anode. These ions pass through positively charged anion exchange membrane,
but are prevented from further migration toward the anode by the negatively charged cation
exchange membrane and therefore stay in the concentrate stream, which becomes concentrated
with the anions. The positively charged species in the dilute stream migrate toward the
negatively charged cathode and pass through the negatively charged cation exchange membrane.
These cations also stay in the concentrate stream, prevented from further migration toward the
cathode by the positively charged anion exchange membrane. Anion and cation migration is
enabled by an electric current that flows between the cathode and anode. A schematic example
of an electrodialysis process is given in Figure 7 for NaCl concentration. Only an equal number of
anion and cation charge equivalents are transferred from the dilute stream into the concentrate
stream, maintaining the charge balance in each stream. The overall result of the electrodialysis
process is an ion concentration increase in the concentrate stream with a depletion of ions in the
dilute solution feed stream (American Water Works Association, 1995).
Figure 7: Example of an electrodialysis process (Wikipedia)
This basic engineering module calculates the energy requirements for the electrodialysis process.
System boundary
The system boundary of this process includes the electrodialysis process in which ions are
shifted from a diluate stream to a concentrate stream. Either the diluate stream is the useful
output (if the aim is to get rid of certain substances in a liquid) or the concentrate stream is the
useful output (if the aim is to concentrate certain substances in a liquid). There is an additional
flow, i.e. the electrode stream. The electrode stream flows past each electrode in the stack. This
stream may consist of the same composition as the feed stream or may be a separate solution
containing different compounds. Depending on the stack configuration, anions and cations from
the electrode stream may be transported into the concentrate stream, or anions and cations from
the diluate stream may be transported into the electrode stream. In each case, this transport is
42
necessary to carry current across the stack and maintain electrically neutral stack solutions. The
system boundary includes pumping requirements for the transport of the streams.
Calculation algorithm
The electricity requirement for electrodialysis is calculated according to (Perry & Green, 1999):
n
FUE
I
ed
In which:
Eed = Energy consumption (J/ mol equivalent salt shifted)
U = Applied voltage (V)
F = Faraday constant (= 96485.34 Amp.s/mol)
I = Current efficiency (dimensionless)
n = number of cell pairs (dimensionless)
The energy consumption is expressed per mol equivalent salt shifted. An equivalent stands for a
unit of charge. This means that, e.g., 1 mol SO42- equals 2 mol equivalents. And likewise, the shift
of 1 mol H2SO4 equals 2 mol equivalents. The meaning of expressing shifts per equivalent appears
from the following formula for current efficiency:
In
FccQ
In
FN iipiI
0
In which:
Ni = Flow of ionic substance i formed from the splitting operation (mol eq/s)
I = Current intensity through the stack (A)
Qp = Volumetric flow of the product (m3/s)
ci = Concentration of ionic substance i in the product stream (mol eq/m3)
ci0 = Concentration of ionic substance i in the inlet stream (mol eq/m3)
The current efficiency basically determines the amount of coulombs shifted per coulomb applied
by means of external power. The amount of coulombs shifted is determined by the charge of the
ions. Each coulomb originates from 1 charge equivalent of an ion. It is represented by the
numerator and expressed in Amp.s/s = Coulomb/s. The amount of Coulombs externally supplied is
determined by the denominator and is expressed in Amp which is also Coulomb/s.
Electrodialysis
System
boundary
Pumpin
g
System
boundary
Concentrate feed
stream (M)
Diluate feed
stream (M)
Concentrate stream
(UO/WE)
Diluate stream
(UO/WE)
Electricity (En) Electrode
stream (A)
43
User and database input
Input values required from user DATAPHYSCHEM DATABUO
Applied voltage
Number of cell pairs
Current intensity through the stack
Volumetric flow of the product
Concentration of ionic substance i in
the product stream
Concentration of ionic substance i in
the inlet stream
Faraday constant -
User output
-
UPR output
From technosphere:
Electricity required, e.g. “Electricity, medium voltage, production UCTE, at grid”.
Example
A small lab-scale electrodialysis process is considered in which a solution of acetic acid is
concentrated. At the start of the process the concentration of acetic acid in the concentrate
stream is equal to that in the diluate stream, i.e. 0.84 mol/l. The process runs for 330 minutes at
a flow rate of 0.174 l/h. At the end of the process, the concentration in the concentrate stream
is 1.0 mol/l. Furthermore the following parameters/ process conditions apply:
Number of cell pairs: 7
Applied voltage: 24 V
Applied current: 0.24 Amp
Faraday constant: 96485.34 (Amp.s)/mol
Molar mass acetic acid: 60.05 g/mol
Inserting the above values in the formulas defined for this process gives a current efficiency of
0.44. The electricity requirement is 0.75 MJ/mol salt shifted, or 3.45 kWh/kg acetic acid or 12.4
MJ/kg acetic acid.
Rule of thumb values
In the BREW study (Patel et al., 2006) literature values for different electrodialysis processes
have been gathered. This yielded a range of energy values, i.e. 0.07-0.34 kWh/mol eq. The
recommended value in this study is to use 0.1 kWh/ mol eq. The respective equivalent value from
the example above is 0.2 kWh/mol. The difference to the generic value according to the BREW
study (factor of 2) may be explained primarily by a rather low current efficiency in the lab-scale
process (44%).
2.2.7 Pressure swing adsorption
Pressure swing adsorption is a technology that is used for the separation of gas species from a
mixture of gases under pressure. It is based on the species molecular characteristics and affinity
for an adsorbent material. Pressure swing adsorption processes rely on the fact that under
pressure, gases tend to be attracted to solid surfaces, or ‘adsorbed’. Special adsorptive materials
are used as a molecular sieve, for example ‘zeolites’. The process then swings to low pressure to
desorb the adsorbent material.
This basic engineering module calculates the energy required to separate a gas from a gas
mixture by pressure swing adsorption.
System boundary
44
The system boundary of this process includes the separation of (a) gas(es) from a gas mixture
by means of pressure swing adsorption. Either removed gas(es) or the remaining gas mixture can
be the useful output product. For this process, energy in the form of electricity is needed as well
as adsorbent material. The adsorbent material can be reused many times however. In the
system boundary, some pumping is required to transport the gases.
Calculation algorithm
An extensive theoretical analysis of a pressure swing adsorption process is given by Huang et al.
(Huang et al., 2008). The power requirements of a pressure swing adsorption process are
calculated according to:
feedfeedbed
atm
feed
feedg curp
pTRP 2
1
11
In which:
P = Power (W)
= Ratio of heat capacities, cp/cv (dimensionless)
Rg = Ideal gas constant, i.e. 8.314 J mol-1K-1
Tfeed = Temperature of the feed (K)
pfeed = Pressure of the feed (bar)
patm = Atmospheric pressure, i.e. 1.013 bar
rbed = Column radius (m)
ufeed = Interstitial gas velocity (m/s)
cfeed = Concentration of the feed stream (mol/m3)
The ratio of the heat capacities for various gases can easily be found in chemical handbooks
(White, 1999; Lange & Dean, 1973) or even on the internet.
The interstitial gas velocity is calculated with:
A
Qu feed
In which:
Q = Volumetric flow rate (m3/s)
Pressure Swing
Adsorption
Pumping
System
boundary
System
boundary
Gas mixture
(M)
Separated gas(es) (UO)
Electricity (En)
Separated gas(es) (WE)
Adsorbent (A)
45
A = Cross sectional area of the bed (m2)
= Void fraction of the bed, i.e. ratio of the void volume to the total volume of the bed
The concentration of the feed stream (all compounds) is calculated based on the ideal gas law:
feedg
feed
feedTR
pc
If the flow rate of the separated gas is known, the total energy requirements (in J/kg) can be
calculated:
v
PE
In which:
E = Energy use (J/kg)
P = Power (W)
v = Flow rate separated gas (kg/s)
User and database input
Input values required from user DATAPHYSCHEM DATABUO
Temperature of the feed
Pressure of the feed
Column radius
Interstitial gas velocity
Concentration of the feed
Flow rate separated gas
Atmospheric pressure
Ratio of heat capacities -
User output
-
UPR output
From technosphere:
Electricity required, e.g. “Electricity, medium voltage, production UCTE, at grid”.
Example
A pilot plant for the purification of hydrogen from a mixture containing 25% H2 and 75% CH4 has
the following dimensions and operating conditions:
Column radius: 0.25 m
Interstitial gas velocity: 0.25 m/s
Temperature of the feed: 303.15 K
Pressure of the feed: 500 kPa
Atmospheric pressure: 101.325 kPa
Ratio of heat capacities: 1.342 (25% H2 and 75% CH4)
Ideal gas constant: 8.31 J/(mol.K)
The concentration of the feed is calculated with the ideal gas law and amounts to 198 mol/m3.
Power requirements are 48.4 kW. Assuming a column radius of 0.5 m and a voidage of 0.4 gives a
flow rate of 0.0196 m3/s. If it were assumed that 25% is hydrogen, the hydrogen flow rate from
the PSA process is 17.7 m3/h. If a hydrogen recovery of 79.6% were assumed and a hydrogen
density of 0.08988 g/l at atmospheric pressure, this results in an energy requirement of 138
MJ/kg hydrogen.
46
2.3 Physical mechanical processes
2.3.1 Mechanical compression – Single stage
Compression uses mechanical energy to increase the pressure of a gas. This can be used for
steam, to recover the latent heat, or for other gases such as air, to obtain higher driving forces.
System boundary
The BUO includes the mechanical energy for the compression in a reciprocating or centrifugal
compressor. It assumes the adiabatic compression of ideal gases, which is a reasonable
approximation for most compressors (Perry & Green, 1999). SUO such as pumping are not
included in the α system boundary.
Compression
α System boundary
Gas (M) Compressed gas (UO)
β System boundary
Pumping
Electricity (En)
Calculation algorithm
The mechanical energy needed for the compression (P, in kW) can be obtained by:
With m the mass flow in kg/s, MM the molar mass and the specific compression work W (in
kJ/kmol). In the case of ideal gases, the specific work can be obtained by (Perry & Green, 1999):
R = universal gas constant (J/molK), Pc = pressure after compression (Pa), pi initial pressure (Pa),
Ti initial temperature (K) = Tb; the boiling temperature, α = efficiency factor, ratio of specific
heats, adiabatic coefficient. For the latter, the values in Table 8 can be used as defaults
(www.engineeringtoolbox.com), however for other gases, or other conditions, this value should
be modified by the user.
Table 8: Default adiabatic coefficients for different ranges
Range Ratio of heat capacities
Steam 0.068 atm 50 – 316 oC 1.32
Steam 1 atm 107 –316 oC 1.31
Steam 10,2 atm 182 – 316 oC 1.28
Air 1.41
47
Temperature and pressure before and after compression can be linked to each other to increase
user friendliness:
With Tc the temperature after compression.
The efficiency factor depends on the type of compressor and the compression ratio or flow. The
user has to choose a type of compressor and default efficiency factors will be taken from
DATABUO (Table 9 (www.cheresources.com)).
Table 9: Default evaporator efficiencies for different types of evaporators
Compression ratio (pc/pi) Efficiency
Reciprocating compressors
1.5 65%
2 75%
3-6 80-85%
Centrifugal compressors
2.8 to 47 m3/s 76-78%
User and database input
Input values required DATABUO DATAPHYSCHEM
Mass and type of feed
Feed temperature
Type of compressor
Compression ratio
Efficiency Adiabatic coefficients
User output
Physical properties of output stream
UPR output
From technosphere:
Electricity use
Example
A vapour leaving an evaporator is recompressed with a compression ratio of 1.5. The intial
temperature is 373K and the adiabatic coefficient is 1.31. The specific work W becomes
2030J/mol or 113kJ/kg and the output temperature becomes 411K.
The UPR α of compressing 1kg of vapour with a compression ratio of 1.5 is thus:
From technosphere
0.031kWh electricity, medium voltage
2.3.2 Mechanical Compression- Multi stage
With multistage compression, the pressure to which gases can be compressed is much higher than
with single stage compression. Another reason for applying multistage compression is that the
same compression task can be realized with lower energy use. Multistage compression is a
sequence of compressions. After each compression stage the heat that is generated in the
compression is removed by cooling, making multistage compression less adiabatic and more
48
isothermal. The choice of the compressor type for each compression step depends mainly on the
flow rate and the differential pressure (Damen, 2007). The theoretical energy requirements can
be estimated, however. This basic engineering module calculates the estimated energy
requirements for multistage compression of gases.
System boundary
The system boundary of this BUO includes the multistage compression of a gas. Electricity is
needed for the compressor. The system boundary includes the pumping and cooling
requirements.
Calculation algorithm
For a multistage compression with n stages, power requirements are given by following equation
(Diomedes Christodoulou, 1984):
11
1
1
21n
p
pn
M
ZRTW
In which:
W = Specific compression work (J/g or kJ/kg)
Z = Compressibility factor (Dimensionless)
R = Universal gas constant i.e. 8.314 J mol-1K-1
T1 = Suction temperature (K)
n = Number of stages (Dimensionless)
= Ratio of heat capacities, cp/cv (dimensionless)
M = Molar mass (g/mol)
p1 = Suction pressure (MPa)
p2 = Discharge pressure (MPa)
Since this is the theoretical power use of the multistage compression process, a correction has to
be made for efficiency losses:
mis
WP
Multistage
compression
System boundary
Pumping Cooling
System boundary
Gas
(M)
Compressed
gas (UO)
Electricity (En)
49
In which:
P = Power requirements (kJ/kg)
is = Isentropic efficiency (Dimensionless)
m = Mechanical efficiency (Dimensionless)
User and database input
Input values required DATAPHYSCHEM DATABUO
Suction temperature
Number of stages
Suction pressure
Discharge pressure
Universal gas constant
Specific heat ratio
Molar mass
Compressibility factor
Isentropic efficiency
Mechanical efficiency
User output
Physical properties of the output stream (pressure, density)
UPR output
From technosphere:
Electricity required, e.g. “Electricity, medium voltage, production UCTE, at grid”.
Example
Koornneef (2010) has assumed multistage compression of carbon dioxide for injection in an
underground well in his study on carbon capture and storage. He has assumed the following
parameters:
Compressibility factor: 0.9942
Universal gas constant: 8.3145 J mol-1 K-1
Suction temperature: 313.15 K
Specific heat ratio: 1.293759
Molar mass: 44.01 g/mol
Suction pressure: 10.7 MPa
Discharge pressure: 15 MPa
Number of compression stages: 2
Isentropic efficiency: 80%
Mechanical efficiency: 99%
When inserting these values in the algorithms defined above, the specific compression work is
20.0 MJ/ton. The specific electricity requirement is 25.2 MJ/ton.
2.3.3 Pumping incompressible fluids
This module includes the pumping of incompressible fluids. Whilst in principle all fluids are
compressible, the compressibility of liquids is low, and for gases, incompressibility is often used
as an approximation at lower speeds.
System boundary
The electricity use is calculated to pump an incompressible fluid over a certain distance and
height. The calculation algorithm can be used for centrifugal, axial, rotary and reciprocating
pumps, for Newtonian and Power law fluids. Herschel – Bulkley can also be chosen, but larger
errors are expected, due to limited knowledge of their behavior in different pipe configurations.
Numerous assumptions are made in the equations: constant fluid density, the absence of thermal
energy effects; single phase, uniform material properties, uniform equivalent pressure (Valentas
et al., 1997).
50
Pump
α System boundary
Incompressible fluid (M) Incompressible fluid (UO)
β System boundary
Electricity (En)
Calculation algorithm
The starting point to estimate the pump power is to determine the type of fluid. Three different
types of fluids are considered depending on the relation between the shear stress (σ) and the
shear rate (γ):
Newtonian fluids:
Power law fluids:
Herschel – Bulkley fluids:
With µ the viscosity, µ’ the plastic viscosity, K the consistency index, n the flow behavior index
and σ0 the yield stress. Often the Newtonian equations are used, but in many real cases such as
different types of slurries the Newtonian theory is not applicable. Even fibrous slurries such as
fermentation broths, fruit pulps, crushed meal animal feed, tomato puree, sewage sludge, and
paper pulp, which may not contain a high percentage of solids may flow as non-Newtonian
regimes (Abulnaga, 2002). Because of its ease of use, the empirical Power Law is often used (Rao,
1999).
Pump power
The basic equation to estimate pump power for transporting incompressible fluids is:
Pn = pump power output in W, H the total dynamic head in Nm/kg, Q the capacity in m³/s, ρ the
density (kg/m³), g the gravity constant and eff. the efficiency of the pump (Perry & Green,
1999).
The main bottleneck for the calculation of the necessary power is thus determining the dynamic
pump head to displace the fluid. This factor depends on the conditions at starting and end
position, on the flow rate, on the type and configuration of the pipes used, and on the type of
fluid that is pumped. The dynamic head can be obtained from the Bernoulli equation: (Valentas
et al., 1997).
With Δz the difference in height between the starting and end position, which equals zero in
closed systems. Δp is the pressure difference between initial and end situation, α is the
correction factor for velocity distribution in the pipe and hw is the resistance head.
Velocity head
The difference in velocity is represented by Δu, where the velocity behind the pump can be
calculated based on the flow rate and the cross sectional area of the pipes (A):
51
According to www.cheresources.com liquid transport pipes should be sized according to:
With D the diameter of the pipe (in m).
For circular pipes:
Filling this in in eq. links the flow (in m³/s) to the diameter (in m):
This can be solved if the flow is known.
α is the correction factor for velocity distribution in the pipe. For the turbulent flow this is
always approximated as 2, whilst for laminar flow it depends on the type of fluid (Ibarz &
Barbosa-Cánovas, 2003):
Newtonian fluids: α=1
Power law fluids (using the flow behavior index n)
Herschel – Bulkley fluids: The correction factor can be deducted from Figure 8:
Figure 8: The kinetic energy correction factor for Herschel-Bulkley fluid foods (Valentas et al., 1997)
52
The dimensionless yield stress (c):
σ0 = the yield stress, σw the shear stress at the wall and ΔP/L the pressure drop per unit of length.
Resistance head
hw is the resistance head (m), which consists out of a basic friction resistance, with f the fanning
friction factor, and a factor kf accounting for supplementary losses, with b the amount of fittings,
valves, elbows,...
This resistance factor depends heavily on the flow regime. Therefore the Reynolds number has to
be determined, which can be calculated for Power Law fluids by:
For fluids without yield stress, the critical Reynolds number is determined with:
When the magnitude of n < 1 the fluid is shear-thinning or pseudoplastic, and when n > 1 the fluid
is shear thickening or dilatant in nature (Rao, 1999). For the special case of a Newtonian fluid (n
= 1), the consistency index K is identically equal to the viscosity of the fluid. Thus for Newtonian
fluids this becomes:
And the critical Reynolds number is 2100, meaning that a Reynolds numbers lower than this value
are laminar. Reynold numbers above 4000 are situated in the turbulent region. For the
intermediate region (2100<Re<4000), it is impossible to obtain the pumping equations, thus an
approximation should be made. Often Reynolds numbers above the critical number (2100 for
Newtonian fluids) are assumed to be in the turbulent region.
When a fluid has a non negligible yield stress, the Herschel – Bulkley theory should be used. The
critical Reynolds number can then be calculated by:
With HeG the Hedstrom number (Ibarz & Barbosa-Cánovas, 2003):
53
The critical Reynolds number can be obtained graphically, however, in this work the Herschel-
Bulkley fluids are only used in the laminar region, which is realistic due to the high viscosity and
elasticity of these fluids.
Fanning friction factor
Laminar flow
Since the Herschel – Bulkley equation is the general form of the three previous equations, this
will be taken as a starting point, where the friction factor can be calculated as:
With
For power law and Newtonian fluids c=0, making Ψ=1 and thus the friction factor can be
simplified to:
For turbulent flow, it is thus assumed that the fluids are obeying Newton, or Power law
equations.
Turbulent flow in smooth pipes:
For Power Law fluids the Dodge and Metzner (1959) equations give good results for (n = 0.4 to 1):
This equation is simplified to the Von Karman equation for Newtonian fluids (n=1) (Chhabra &
Richardson, 1999):
For turbulent flow in rough pipes, the roughness of the pipe has an influence. However, for non
Newtonian flows, these relationships are not well studied.
The Torrance equation is used for fluids with n<1 (Liu, 2003).
D represents the inner diameter and ε the absolute roughness which depends on the material
used. Default values can be found in DATABUO (Table 10) (Van Der Meeren, 2004). For other fluids
it is stated that the equations valid for Newtonian fluids can be used as an approximation, since
54
the turbulence becomes more important (Abulnaga, 2002). A popular equation for Newtonian
fluids is given by the Colebrook-White equation (Perry & Green, 1999):
Table 10: Default roughness factors of different piping material
Type Absolute roughness ε in m
PVC, plastic, glass 0
Drawn tubing 0.0000015
Commercial steel and wrought iron 0.000045
Asphalted cast iron 0.00012
Galvanized iron 0.00015
Cast iron 0.00026
Wood stave 0.00018-0.00092
Concrete 0.00030-0.0030
Riveted steel 0.00092-0.0092
Supplementary losses
The second term in friction factor accounts for supplementary losses. In general these factors are
determined experimentally for Newtonian fluids and due to insufficient data they are also used
for non-Newtonian fluids (Perry & Green, 1999; Chhabra & Richardson, 1999). In general two
options exist. For elbows, gates, valves, equivalent lengths are used:
With equivalent lengths (in m) taken from Table 11 (Brannan, 2002):
55
Table 11: Equivalent lengths for different piping parts
Nom
inal pip
e d
iam
ete
r cm
Glo
be
valv
e
or
ball
check
valv
e
Angle
valv
e
Sw
ing c
heck v
alv
e
Plu
g c
ock
Gate
or
bal valv
e 45° e
llbow
Short
ra
diu
s
elb
ow
Long
radiu
s
elb
ow
Hard
T
Soft
T
90
mit
er
bends
Weld
ed
Thre
aded
Weld
ed
Thre
aded
Weld
ed
Thre
aded
Weld
ed
Thre
aded
Weld
ed
Thre
aded
2 m
iter
3 m
iter
4 m
iter
3.81 16.76 7.92 3.96 2.13 0.30 0.30 0.61 0.91 1.52 0.61 0.91 2.44 2.74 0.61 0.91
5.08 21.34 10.06 5.18 4.27 0.61 0.61 0.91 1.22 1.52 0.91 1.22 3.05 3.35 0.91 1.22
6.35 24.38 12.19 6.10 3.35 0.61 0.61 - 1.52 - 0.91 - 3.66 0.91 -
7.62 30.48 15.24 7.62 5.18 0.61 0.61 1.83 1.22 4.27 1.22
10.16 39.62 19.81 9.75 9.14 0.91 0.91 2.13 1.52 5.79 1.52
15.24 60.96 30.48 14.63 21.34 1.22 1.22 3.35 2.44 8.53 2.44
20.32 79.25 38.10 19.51 36.58 1.83 1.83 4.57 2.74 11.28 2.74
25.4 100.58 48.77 24.38 51.82 2.13 2.13 5.49 3.66 14.33 3.66
30.48 121.92 57.91 28.96 51.82 2.74 2.74 6.71 4.27 16.76 4.27 8.53 6.40 6.10
35.56 137.16 64.01 32.00 24.38 3.05 3.05 7.92 4.88 18.90 4.88 9.75 7.32 6.71
40.64 152.40 73.15 36.58 44.20 3.35 3.35 8.84 5.49 21.95 5.49 11.58 8.23 7.32
45.72 167.64 85.34 42.67 48.77 3.66 3.66 10.06 6.10 24.99 6.10 12.80 9.14 8.53
50.8 198.12 91.44 47.24 64.01 4.27 4.27 10.97 7.01 27.43 7.01 14.02 10.06 9.75
55.88 209.70 102.11 51.82 68.58 4.57 4.57 12.19 7.62 30.48 7.62 15.85 10.97 10.36
60.96 228.60 112.78 56.39 77.42 4.88 4.88 13.41 8.23 33.53 8.23 17.07 11.89 10.97
76.2 95.10 6.40 6.40 16.76 12.19 42.67 12.19 21.34 15.54 13.41
91.44 7.62 7.62 20.12 14.33 51.82 14.33 25.60 18.29 15.85
106.68 9.14 9.14 23.47 16.76 60.96 16.76 29.87 21.03 19.51
121.92 10.67 10.67 26.82 19.81 67.06 19.81 34.14 24.69 21.95
137.16 12.19 12.19 30.18 21.34 76.20 21.34 38.40 27.43 24.38
152.4 13.72 13.72 33.53 24.38 79.25 24.38 57.91 30.18 28.04
For enlargements and narrowings, a dimensionless coefficient is used:
With kf a dimensionless loss coefficient, which depends on several parameters for different
occasions (Van Der Meeren, 2004):
From pipe to reservoir Kinetic energy gets lost = 1 independent of geometry
From reservoir to pipe
o Well rounded inlet kf = 0.04
o Chamfered inlet kf =0.25
o Square edge inlet kf =0.5
o Inward projecting pipe kf =1
For abrupt pipe enlargement (d1<d2)
56
For gradual pipe enlargements values from Table 12 can be chosen.
Table 12: Dimensionless loss coefficient for gradual pipe enlargements
D2/D1
Angle of cone °
2 6 10 15 20 25 30 35 40 45 50 60
1.1 0.01 0.01 0.03 0.05 0.10 0.13 0.16 0.18 0.19 0.20 0.21 0.23
1.2 0.02 0.02 0.04 0.09 0.16 0.21 0.25 0.29 0.31 0.33 0.35 0.37
1.4 0.02 0.03 0.06 0.12 0.23 0.30 0.36 0.41 0.44 0.47 0.50 0.53
1.6 0.03 0.04 0.07 0.14 0.26 0.35 0.42 0.47 0.51 0.54 0.57 0.61
1.8 0.03 0.04 0.07 0.15 0.28 0.37 0.44 0.50 0.54 0.58 0.61 0.65
2.0 0.03 0.04 0.07 0.16 0.29 0.38 0.46 0.52 0.56 0.60 0.63 0.68
2.5 0.03 0.04 0.08 0.16 0.30 0.39 0.48 0.54 0.58 0.62 0.65 0.70
3.0 0.03 0.04 0.08 0.16 0.31 0.40 0.48 0.55 0.59 0.63 0.66 0.71
∞ 0.03 0.05 0.08 0.16 0.31 0.40 0.49 0.56 0.6 064 0.67 0.72
Abrupt pipe narrowing (d1>d2)
For non cylindrical pipes
D = De with
Efficiency
The pump efficiency (Eff.) can be calculated by (Brannan, 2002):
With F the developed head (in ft) and G the flow (in GPM). To convert to SI units (head H in
meters and flow rate Q in m³/s):
The equation is valid for F=50-300ft and G=100 – 1000GPM
If this is not the case, default values can be used (Table 13) (www.cheresources.com):
57
Table 13: Default pump efficiencies for pumping incompressible fluids
Pump type flow m³/s Efficiency
Centrifugal pump
0.0063 45%
0.0315 70%
0.63 80%
Axial pump all 75%
Rotary pump all 65%
Power (kW)
Reciprocating pump
7.46 70%
37.3 85%
373 90%
User and database input
Input values required DATABUO DATAPHYSCHEM
Mass (flow rate) and type of fluid
Starting and end temperature and pressure
Type pipe material, distance of straight
pipes, height difference between starting
and end point, number and type of turns,
valves, enlargements and narrowings
Friction coefficients
and roughness factors
Efficiencies
Fluid characterisation factors
User output
-
UPR output
From technosphere:
Electricity use
Example
A Newtonian fluid is pumped at a rate of 2m³/s. Total height difference between starting and end
point is 3m and total pipe length is 20m. Starting and end pressure are equal and the end velocity
is zero. If the pipe is made of riveted steel and if the fluid is pumped via 2 valves, 2 short radius
elbows and 2 long radius elbows, the power use of a rotary pump is 441 kWh/h.
The UPR of pimping 1m³ fluid over this distance becomes:
From technosphere
0.061kWh electricity, medium voltage
2.3.4 Pumping incompressible fluids through packing
This module includes the pumping of incompressible fluids through packings, beds or filters. It
can thus be used for pumping operations such as adsorption, filtration, catalytic beds, etc.
However, only the electricity use is modeled of the pumping operation. No interactions between
the fluid and the bed is included.
System boundary
The calculations in this module can be widely applied, for several unit operations where beds and
packings are involved. However, the equations are only valid if the characteristics of the bed
remain constant and if no fluidization occurs. Especially the porosity should be constant (Van Der
Meeren, 2004). For example clogging filters are out of the range of this work. The module can be
58
coupled to the pumping operations in 4.3.2. since they only include the pumping through the
bed. The same pumps and pump efficiencies are included as for the regular pumping module.
Pumping through packing
α System boundary
Incompressible fluid (M) Incompressible fluid (UO)
β System boundary
Electricity (En)
Calculation algorithm
The approach is similar to pumping incompressible fluids. However, on top of the regular friction
term, an additional head loss is caused by the packing material. To account for this, the approach
from Chhabra & Richardson (1999) can be used where the bed head loss is added to the modified
Bernoulli equation:
Which can be used in:
The head loss of the bed (in m) can be calculated from the pressure drop:
With:
With ε the pore volume/bed volume, u the superficial velocity (m/s), L the length of the bed (m),
ρ the density of the fluid (kg/m³), dp is the sphere diameter (m). The latter depends strongly on
the particle shape:
With Г the sphericity of the particles and dr the equal volume sphere diameter (Table 14) (Ibarz &
Barbosa-Cánovas, 2003; McCabe et al., 2004)
59
Table 14: Sphericity of different packing particles
Form of the particle Sphericity (Г)
Sphere 1
Cube 0.81
Cylinders
h=d
h=5d
h=10d
0.87
0.70
0.58
Discs
h=d/3
h=d/6
h=d/10
0.76
0.60
0.47
Beach sand As high as 0.86
River sand As low as 0.53
Other types of sand 0.75
Triturated solids 0.5-0.7
Granulated particles 0.7-0.8
Wheat 0.85
Raschig rings 0.26-0.53
Berl saddles 0.3-0.37
Coal dust 0.73
Mica flakes 0.28
Crushed glass 0.65
Cf is a correction factor to account for the fact that the packing material is more dense in the
center of the column and less dense near the wall. It can be calculated for cylindrical vessels:
With D the vessel diameter
fbed is the friction factor of the bed, which can be obtained from the Ergun equation:
With the Reynolds number for Power law fluids:
It is stated that this equation is a good approximation for ε ≤ 0.41 and Re*<100.
For ε > 0.41 and Re*>100
With
60
User and database input
Input values required DATABUO DATAPHYSCHEM
Mass (flow rate) and type of fluid
Type and size of packing material
Pore volume/bed volume (should remain
constant)
Sphericity
Pump efficiency Fluid characterisation factors
User output
-
UPR output
From technosphere:
Electricity use
Example
Pumping 0.5m³/s water through a vertical packed bed of 2 m length with mica flakes of 1cm and
a porosity of 0.4 requires a rotary pump shaft power (65% efficiency) of 268kW.
The UPR α of pumping 1m³ water through this bed is thus:
From technosphere
0.15 kWh electricity, medium voltage
2.3.5 Agitation and mixing of liquids and suspensions
Agitation and mixing are operations often needed in process industry and relying on a certain
amount of mechanical energy which depends on many different parameters. This module includes
mixing two or more liquids or suspensions and agitation of one or more liquids or suspensions.
Whilst these two operations might not exactly have the same purpose, their power consumption
can be modelled with the same equations.
System boundary
The mixing and homogenization of liquids operation includes the electricity use necessary for the
impeller. Several types of propellers and turbines are included, plus a paddle with 2 blades, an
anchor and a helical impeller. Equations are available for Newtonian and Power Law fluids, the
latter thus also including pseudoplastic and dilatants suspensions. Furthermore, the effect of gas
bubbling can be included. However, the power for the gas bubbling and pumping operations are
SUO in the β system boundary.
61
Agitation and mixing
α System boundary
Feed (M) Agitated or mixed feed (UO)
β System boundary
Pumping
Electricity (En)
Bubbling
Calculation algorithm
Newtonian fluids
One of the most important factor for the power requirement is the Reynolds number. Similarly to
pumping fluids, a difference can be made between Newtonian and Non-Newtonian fluids. In the
latter case the main complication is related to the shear rate (γ). The calculation in the two
situations is very similar, starting from the Reynolds number (McCabe et al., 2004):
With N the rotational speed (rps), DA the impeller diameter (m), µa the apparent viscosity (Pa.s)
and ρ the density of the fluid (kg/m³).
Newtonian fluids
For Newtonian fluids: µa = µ
The standard equation for the total power needed for mixing and homogenization (P in W) is
calculated by:
This is valid in for intermediate Reynolds numbers between 100<NRe<10000, whilst for the
turbulent region, NRe>10000, the modified version is used:
In the laminar region the viscosity and shear become more important, thus the formula is
converted to:
Power numbers for different flow regimes and impellers can be found in Table 15.
62
Table 15: Power numbers for newtonian fluids in the laminar (KL), intermediate (P0) and turbulent region (KT) for different types of impellers (McCabe et al., 2004)
typical
viscosity (cp) Type KL P0 KT
<2000 Average small propeller (3 blades) 41.00 0.75 0.32
<2000 Average small propeller with pitch of 2 (3
blades) 43.50 0.75 1.00
<2000 Average large propeller (3 blades) 41.00 0.75 0.32
<2000 Average large propeller with pitch of 2 (3
blades) 43.50 0.75 1.00
<20000 Turbine 6 flat blades 71.00 5.00 6.30
<20000 Turbine 6 curved blades 70.00 5.00 4.80
<20000 Fan turbine 6 blades 70.00 5.00 1.65
<80000 Paddle (2 blades) 36.50 2.60 1.70
<100000 Anchor 300.00 10.00 0.35
<1000000 Helical impeller 300.00 10.00 0.35
>1000000 Extruders. roll mill. etc
Non Newtonian fluids
For non Newtonian fluids, the apparent viscosity should be used in the Reynolds number
(Heldman & Lund, 2007):
For dilatants liquids (Perry & Green, 1999)
For pseudoplastic and Bingham fluids
For turbines 10 can be replaced by 11.5.
The Power Number can then be found in Figure 9 (Heldman & Lund, 2007) for different types of
impellers and it can be applied in:
63
Figure 9: Power number correlations
Vessel design
The diameter of the impeller is estimated from the volume of the cylindrical tank, on its turn
calculated from the flow (Q in m³/s), and the residence time (t in s):
A certain excess volume is added (Ex. in %) to add a buffer volume:
Standard this excess volume is set on 10%.
A cylindrical vessel is assumed with H = Dt with H the height of the tank and Dt the diameter of
the tank:
64
Furthermore, we assume that the diameter of the impeller is 1/3rd of the tank diameter:
Rotational speed and mixing time
For mixing operations, the residence time and impeller speed are 2 interlinked parameters which
therefore cannot be fixed by default values. According to Herbert et al. 1994 (a dimensionless
mixing time Θ can link the mixing time (t) with the impeller speed for baffled agitated vessels
with a centrally located impeller:
With NP the power number. Typical N values for the different impellers are presented in Table 16
and typical power numbers can be found in Table 15 for laminar, turbulent and intermediate
region.
Table 16: typical rotational speeds for different impellers in mixing vessels (McCabe et al., 2004)
Type Typical N ranges for mixing
Average small propeller (3 blades) 1150-1750
Average small propeller with pitch of 2 (3 blades) 1150-1751
Average large propeller (3 blades) 400-800
Average large propeller with pitch of 2 (3 blades) 400-801
Turbine 6 flat blades 200-400
Turbine 6 curved blades 200-400
Fan turbine 6 blades 200-400
Paddle (2 blades) 20-150
Anchor 50-350
Helical impeller 5-20
The rotational speed of disc turbine blades applied in Newtonian fluids can be estimated by using
a scale of agitation SA based on the pumping number (NQ) which can be found in Table 17
(Chhabra & Richardson, 1999):
The scale of agitation ranges from 1, which is mildly mixed, to 10, which can be taken for intense
mixing.
65
Table 17: Default pumping numbers for types of impellers
Type Pumping number (NQ)
Average small propeller (3 blades) 0.5
Average small propeller with pitch of 2 (3 blades) 0.5
Average large propeller (3 blades) 0.5
Average large propeller with pitch of 2 (3 blades) 0.5
Turbine 6 flat blades 0.7
Turbine 6 curved blades 0.8
Fan turbine 6 blades 0.8
Paddle (2 blades) 0.6
Anchor 0.5
Helical impeller 0.5
For gassed liquids, the power (Pg) for mixing and homogenization can be calculated by:
With σ the surface tension (N/m) of the liquid and NQ the pumping number.
Alternatively an adjusted power number (Np.g) can be calculated:
With Qt the volumetric gas flow rate (kg/s).
User and database input
Input values required DATABUO DATAPHYSCHEM
Mass flow and type of feed(s)
Type of impeller
Impeller speed and residence
time
Tank volume
Power number and pumping
number
Impeller speed and residence
time information
Physicochemical properties of
mixture
User output
Tank volume
Residence time
UPR output
From technosphere:
Electricity use for the impeller
Example
Considering a fermentation tank with a total volume of 54.2m³ (with an excess volume of 10%).
The used diameter of the impeller is thus 1.37m. When taking the viscosity and density of water
as an approximate, and an impeller speed of 10 rpm for a default fermentation, the Reynolds
number is in the turbulent region (374908). For a turbine with 6 flat blades the power number is
6.3, meaning that the total power required becomes 143kWh/h.
66
The UPR α of the agitation of 1m³ fermentation medium during 1 hour is:
From technosphere:
2.93kWh electricity, medium voltage
Rule of thumb values
In the BREW study (Patel et al., 2006) literature values for agitation have been collected ranging
from 0.1-12 kW/m3 of medium. The recommended value was determined at 0.5 kW/m3
2.3.6 Comminution
Comminution is a process which is applied to reduce the size of solids and is widely used in
industry (e.g. in food processing, minerals processing, ceramic industry and so on). The purposes
of comminution are to liberate certain compounds for concentration processes, to reduce the
size or to increase the surface area. More recent technologies result in the need to modify the
surface of solids, prepare composite materials and to recycle the useful components of industrial
waste. The energy efficiency of comminution is very low and the energy required for
comminution increases with a decrease in feed or produced particle size. In design, operation
and control of comminution processes, it is necessary to correctly evaluate the comminution
energy of solid materials (Kanda & Kotake, 2007). In this basic engineering module, the energy
for comminution is determined.
System boundary
The system boundary of this BUO includes the comminution of a feed in to a grinded product.
Electricity is required for this process. No additional processes are assumed in the system
boundary.
Calculation algorithm
Several theories have been developed to estimate the energy requirement of size reduction
processes. They are in fact all based on the basic assumption that the energy required to produce
a change dL in a particle of typical size dimension L is a simple power function of L:
nLKdLdE /
Where dE is the differential energy required, dL is the change in a typical dimension, L is the
magnitude of a typical length dimension and K and n are constants.
The most widely applied are the theories from Kick, Rittinger and Bond. We will now discuss each
of them (Earle & Earle, 2004).
Kick’s law
Comminution Feed (M)
System boundary
System boundary
Grinded product
(UO)
Electricity (En)
67
Kick assumed that the energy required to reduce a material in size is related to the change in
particle diameter, i.e. to the ratio of the diameter of the particles before and after
comminution. In this case ‘n’ in the above equation is -1. The following formula is known as
‘Kick’s law’:
2
1lnL
LfKE cK
In which:
E = Energy for size reduction (J/kg)
KK = Kick’s constant (m3/kg)
fc = Compressive strength of the material (N/m3)
L1 = Diameter of the feed particles (m)
L2 = Diameter of the product particles (m)
Kick’s law is mainly used for comminution of coarse particles.
Rittinger’s law
Rittinger, on the other hand, assumed that the energy for size reduction is directly proportional
to the change in surface area and not to the change in length dimensions. This means that ‘n’ in
the above equation is -2, resulting in ‘Rittinger’s law’:
12
11
LLfKE cR
In which:
KR = Rittinger’s constant (m4/kg)
Rittinger’s law is mainly used for comminution of fine particles.
Bond’s law
Bond has suggested an intermediate course, in which he postulates that ‘n’ is -3/2. This leads to:
qLEE i
11
100
2
and
2
1
L
Lq
In which:
E = Energy for size reduction (kWh/tonne)
Ei = Work index (kWh/tonne)
L1 = Feed particle size ( m)
L2 = Product particle size ( m)
Bond defines work index Ei as the amount of energy required to reduce a unit of mass of the
material from an infinitely large particle size down to a particle size of 80% passing 100 m (Note
that therefore L1 and L2 have to be expressed in m as well!). Table 18 gives an overview of
typical values for the work index for various materials (Perry & Green, 1999).
Table 18: Work Index for various materials in kWh/tonne (Perry & Green, 1999)
68
In literature, data for Bond’s work index are readily available, while this is not the case for
Rittinger’s and Kick’s constants. Since Bond’s law is known to be a general law that is
intermediate to Rittinger’s and Kick’s law, we propose to use Bond’s law in comminution
calculations.
User and database input
Input values required DATAPHYSCHEM DATABUO
Feed particle size
Product particle size
Compressive strength of the material
Rittinger’s constant of the material
Kick’s constant of the material
Work index of the material
-
User output
-
UPR output
From technosphere:
Electricity required, e.g. “Electricity, medium voltage, production UCTE, at grid”.
Example
In the first example, we assume Kick’s law for grinding particles of an unspecified substance from
45 mm down to 4 mm. The compressive strength of this material is 22.5 N/m2 and Kick’s constant
is 239 m3/kg. The energy use for size reduction, hence, is 13.0 kJ/kg.
In the second example we will assume Rittinger’s law for grinding limestone particles with a
diameter of 6 mm down to a size of 0.4 mm. The compressive strength of this material is 70 N/m2
and Rittinger’s constant is 0.013 m4/kg (Backhurst & Harker, 2002). Using Rittinger’s law, the
energy requirement for comminution is 8.95 kJ/kg.
69
In the third example, we assume Bond’s law for crushing the limestone from the previous
example. The work index for limestone is 11.61 kWh/tonne (See Table 18). Applying these values
in the equation for Bond’s law, yields an energy use of 4.3 kWh/tonne or 15.5 kJ/kg.
2.3.7 Fluidization
Fluidization occurs when a bed of granular material is converted from solid to a fluid or
suspended state through the velocity of a gas or liquid. It thus occurs at a certain velocity, where
a minimum fluidization velocity is necessary to achieve the fluidized state, higher velocities will
result in the transport of the particles, and thus in (pneumatic in the case of a gas) conveying.
Fluidization is often used in the process industries during cracking, toasting, roasting( pyrite.
lime. coffee), drying (grains, sugars), freezing, heating with sand baths, encapsulation and
agglomeration of particles, etc. (Van Der Meeren, 2004).
System boundary
This BUO includes the mechanical energy (electricity) to obtain a fluidized state. The relationship
is based on the pressure drop in the bed and is thus valid for fluidization and conveying the
solids. However, it only accounts for obtaining a static fluidized state. To really lift and move the
solids, as is the case in conveying, additional energy is required which is supplied by the fluid.
This additional demand should be calculated with the SUO ‘pumping’.
Fluidization
α System boundary
Fluid (M)
Fluidized bed (UO)
β System boundary
Pumping
Electricity (En)
Solid particles (M)
Calculation algorithm
The basic equation to estimate pump power for fluidization and pneumatic conveying is (Van Der
Meeren, 2004):
With Pn = pump power output in W, Q the capacity in m³/s.
The pressure drop (Pa) over the bed can be calculated by (Richardson, Harker, & Backhurst,
2002):
ρs is the density of the particles and ρf is the density of the fluid (kg/m³). L is the length of the
bed (m), g is the gravity constant (9.81m/s²), whilst the value of ε represents the porosity. If this
70
value is not known, the user can calculate the minimum porosity for fluidization where ε = εmf
(Ibarz & Barbosa-Cánovas, 2003):
With dp the diameter of the particles in µm. This equation is valid for particle sizes going from 50
to 500µm.
User and database input
Input values required DATABUO DATAPHYSCHEM
Mass flow and type of feed(s)
Diameter of particles
- Densities
User output
-
UPR output
From technosphere:
Electricity use
Example
A water flow of 1m³/s holds a fluidized bed of m length with particles of 500µm and a density of
1500kg/m³. The minimal porosity becomes 40%. The pressure drop is 5886 Pa and the required
power 5.886kW.
The UPR α of maintaining the solids in a fluidized state by water for 1 h is thus:
From technosphere
5.886kWh electricity, medium voltage
2.3.8 Conveying solids
Conveying is a frequently used application to transport solids.
System boundary
This BUO includes the electricity to move solids over a certain distance with 4 types of conveyors:
Screw conveyors
Belt conveyors
Centrifugal-discharge buckets on belt
Continuous buckets on a chain
71
Conveying
α System boundary
Solids (M) Solids (UO)
β System boundary
Electricity (En)
Calculation algorithm
Conveying solids requires a certain amount of power which can be found in the Tables 21.6-21.9
(Perry & Green, 1999) for different types of conveyors for certain capacity flow rates.
72
73
User and database input
Input values required DATABUO DATAPHYSCHEM
Mass flow
Type of conveyor Power of equipment -
User output
-
UPR output
From technosphere:
Electricity used
Example
10 tons/h wood chips are transported with a screw conveyor over a distance of 23 m. This
operation needs a motor of 2.76kW.
The UPR of conveying 1kg wood chips is:
From technosphere
0.000276kWh electricity, medium voltage
2.3.9 Fans, Blowers & vacuüm pumps
Fans and blowers are used to create a driving force to supply a gas, usually air. Examples of
applications are ventilation and air blowing in a combustion chamber. A vacuum pump works
oppositely, by lowering the pressure.
System boundary
This BUO includes the electricity use of a fan or blower used to increase the pressure of the gas
or the electricity of a vacuum pump, to remove a gas from a certain space.
74
Fan / Blower / Vacuum pump
α System boundary
Gas (M) Gas (UO)
β System boundary
Electricity (En)
Calculation algorithm
The power (P in W) needed can be calculated by (Perry & Green, 1999):
Where Q is the flow rate (m³/s), ΔP is the pressure difference in Pa, and eff is the efficiency of
the device.
Typical pressure differences for vacuum pumps, fans and blowers can be found in Table 19 (Perry
& Green, 1999; Brannan, 2002; Silla, 2003), whilst typical efficiencies can be found in Table 20
(Bureau of Energy Efficiency India).
Table 19: Typical pressure differences covered in vacuum pumps, fans and blowers
Type of device Typical pressure differences (Pa)
Vacuum pump 100000
Fans 7500
Blowers 50000
Table 20: Default efficiencies
Type of fan Efficiency
Centrifugal fan/blower
Airfoil, backward curved/inclined 81
Modified radial 75
Radial 72
Pressure blower 63
Forward curved 62
Axial fan
Vane axial 81
Tube axial 69
Propeller 47
Other
Other 80
User and database input
Input values required DATABUO DATAPHYSCHEM
Mass flow
Type of device
Typical pressure differences
Efficiencies -
75
User output
-
UPR output
From technosphere:
Electricity used
Example
A radial centrifugal fan (72% efficiency) used for the ventilation of a room displaces an air stream
of 1m³/s with a pressure increase of 7500 Pa. The power of the equipment is 10.4kW.
The UPR α for pumping 1m³/s is thus:
From technosphere:
0,0029kWh electricity, medium voltage
2.4 Utilities
2.4.1 Heating
In many industrial processes, liquid substances are heated to a certain temperature and kept at
that temperature for a given amount of time. These liquids can be pure chemicals, mixtures,
solutions or suspensions. Energy is needed to heat the liquid to the required temperature but also
to compensate for heat losses through the wall of the vessel in which it is heated. The latter is
specifically relevant when a certain temperature has to be retained for a longer period. In this
basic engineering module, we will estimate the energy required for heating a liquid in a vessel
for a given amount of time.
System boundary
The α system boundary for this process includes heating the liquid in a vessel. If, pumping or
mixing are required, the energy use can be estimated with the algorithms of those processes as
defined in the respective basic engineering modules for agitation or pumping. The β system
boundary includes the production of the energy carrier that is needed to heat the liquid. This
could for example be steam or natural gas.
Calculation algorithm
For this process, energy is needed for heating the liquid up to a required temperature and
maintaining this temperature for a certain amount of time. The energy needed for heating to a
certain temperature depends on the specific heat of the liquid and can be calculated with:
Heating liquid in
Vessel
System boundary
System boundary
Heating
(En)
Feed
(M)
Heated feed
(UO)
Energy
carrier
76
0TTmcQ eph
In which:
Qh = Energy needed for heating (J)
cp = Specific heat of the liquid (J kg-1K-1)
m = mass of the liquid (kg)
T0 = Initial temperature of the liquid
Te = End temperature of the liquid
Vessels will have walls that are insulated with isolation material (e.g. polyurethane) to prevent
the loss of heat through the walls. However, insulation will never be 100% and some heat is lost
resulting in a decrease of the temperature of the liquid. Extra heating is required to compensate
for this heat loss. The required heat is determined with (Blok, 2007):
d
tATQr
In which:
Qr = Energy required for remaining the temperature at Te (J)
λ = Thermal conductivity of vessel wall material (W m-1K-1)
ΔT = Temperature difference across the wall (K)
A = Surface area of the vessel wall (m2)
t = Period of remaining temperature at Te (s)
d = Thickness of the vessel wall (m)
The heat required for the two processes just described is the theoretical minimum amount of
energy needed. There will be an efficiency loss from burning a fuel to transmission of heat. In
order to calculate the total amount of energy needed (Qt), we have to account for the heating
efficiency (η) :
rht
QQQ
User and database input
Input values required from user DATAPHYSCHEM DATABUO
Mass of the liquid Specific heat of the liquid Heating efficiency
End temperature of the liquid Thermal conductivity of vessel wall
material
Initial temperature of the liquid
Surface area of the vessel wall
Thickness of the vessel wall
Period of remaining Te
Temperature difference across the vessel
wall
User output
Heating energy required (in J) for a given amount of liquid
UPR output
From technosphere:
Amount of heat from, e.g., natural gas. For example: “Heat, natural gas, at boiler
atmospheric non-modulating < 100kW”.
77
Example
We consider the following situation; Water is heated in a vessel from room temperature up to 80 oC. It is kept at this temperature for 1 hour. The vessel has the dimensions H = D = 1 m. A
polyurethane layer with a thickness of 10 cm insulates the vessel. In this situation the following
parameter values apply:
Specific heat of water: 4.18 kJ/(kg.K)
Difference initial (20 oC) and final temperature: 60 K
Temperature difference across the vessel wall: 60 K
Mass of the water: 783.8 kg
Thermal conductivity of polyurethane: 0.014 (W m-1K-1)
Thickness of the vessel wall: 0.1 m
Surface area of the vessel walls: 4.71 m2
Period of remaining the temperature: 3600
With these parameter values the required heat for heating up the water (Qh) is 197 MJ. The heat
required to keep the temperature at 80 oC for one hour (Qr) is 0.143 MJ. If a heating efficiency of
95% is taken into account, the total amount of fuel needed (Qt) is 207.1 MJ.
2.4.2 Cooling
System boundary
Cooling calculates the amount of cooling medium required to cool/condensate a certain feed
stream. The electricity use for pumping the cooling medium should be added with the SUO
‘pumping’. This is not a separation process, thus in the case of a mixture, all components are
assumed to end in the same phase at the same temperature and pressure.
Cooling
α System boundary
Feed (M) Cooled feed (UO)
β System boundary
Pumping
Calculation algorithm
The needed cooling energy (q) can be calculated with:
With λ the latent heat of the condensed feed stream, cp the mean heat capacity of the feed in
gas (g) and liquid (l) phase , Tb the boiling temperature of the mixture, Tf the temperature of the
feed, and Te the end temperature.
The amount of cooling medium can then be calculated by:
With cp the heat capacity of the cooling medium and ΔT the temperature difference of the
cooling medium before and after use.
78
User and database input
Input values required DATABUO DATAPHYSCHEM
Mass flow
In and output temperature -
Boiling temperature
Heat capacities and latent heats
User output
Amount of cooling medium required
UPR output
Electricity use is calculated with the SUO ‘pumping incompressible fluids’.
Example
Cooling a water stream of 1kg/s from 90 to 20 degrees requires 0,034 kg/s refrigerant R134a
which is heated 10°C.
2.4.3 Steam generation
Steam is used in many chemical processes as a heat source. To produce steam, water is
evaporated with heat coming from the incineration of a fuel such as natural gas. In this basic
engineering module we will determine the heat requirements for the production of steam and the
energy that can be obtained when using steam as heat medium.
System boundary
The system boundary of this BUO includes the evaporation of water by means of an externally
supplied heat source. The water is pre-treated with chemicals in order to remove impurities and
hence avoid foaming and scaling, resulting in lower boiler efficiency, more maintenance, lower
boiler life and other problems (Spirax-Sarco Limited, 2001). The required chemicals have not
been taken into account in the system boundary but are part of the system boundary. This
also holds for the pumping requirements for returning condensate.
Calculation algorithm
The energy for heating water to its evaporation temperature, for evaporation of water as well as
the energy required to further heat steam depends on the pressure of the liquid and the steam.
The pressure at which a liquid at a certain temperature evaporates is called the ‘saturated
pressure’. In this basic engineering module, we will assume ‘saturated steam (i.e. at the boiling
temperature at saturated pressure)’ as a default.
Steam used for industrial purposes will mostly be applied in closed systems. Figure 10 is a
schematic representation of such a system; water is evaporated to steam in a boiler at a certain
Steam
generation
System boundary
Water
(M)
Heat (En)
Steam
(UO)
System boundary
Condensat
e pumping Chemical
additives
Feedwater
pretreatm
ent
79
pressure and temperature, the steam is transported to the process in which it is needed, it
condenses back to water transferring heat to the process and the condensate is returned to the
boiler where it is again evaporated to steam.
Figure 10: Closed steam system
In such a system, the water that is evaporated to steam comes from the returned condensate
that has a heat content of hf in case no heat losses are assumed. With heat losses, however, the
enthalpy of the water arriving at the boiler is referred to as h0 (see algorithms below). In the
boiler it is evaporated to steam with a heat content of hg. The energy needed for generating and
delivering a kilogram of steam (useful heat, i.e. 100% of the heat directly used in a process) is
the enthalpy of evaporation (hfg) at a certain temperature and saturated pressure, corrected for
the efficiency of the boiler (η1) and the efficiency of steam distribution and heat transfer (η2).
The amount of energy that can be obtained from the steam is equal to the condensation enthalpy
(which is identical with the evaporation enthalpy):
In which:
Eg = Energy required for steam generation (kJ)
m = Mass of the steam (kg)
hfg = Enthalpy of evaporation (kJ/kg)
η1 = Boiler efficiency (dimensionless, generally around 0.9)
η2 = Efficiency of steam distribution and heat transfer (dimensionless, generally around 0.9)
The values for hfg, hf and hg are obtained from steam tables, published by thermodynamic
handbooks. Table 21 gives an overview of enthalpies at different temperature- and saturated
pressure levels.
In case no closed system is assumed and water has to be heated from room temperature to the
evaporation temperature or in case heat losses in the returned condensate are taken into
account, the algorithm is as follows:
In which:
Ef = Energy needed for heating the water (kJ)
m = Mass of the water (kg)
hf = Enthalpy of the saturated water (kJ/kg)
Boiler Process requiring
heat
Steam with enthalpy hg (= hf+hfg)
Condensate with
enthalpy hf
Evaporation
enthalpy
(hfg)
Condensation
enthalpy (- hfg)
Efficiency boiler = Conversion efficiency
including heat losses in
transportation system =
x
80
h0 = Enthalpy of the water arriving at boiler (kJ/kg)
The value of h0 depends on the pressure and temperature of the water as it arrives at the boiler.
It can be determined with the help of a steam calculator. See for example:
http://www.steamtablesonline.com/steam97web.aspx
For water at atmospheric pressure and a temperature of 20 oC, the specific enthalpy (h0) is 84.01
kJ/kg. In some cases, returned condensate is expanded to lower pressure to generate flash
steam. This flash steam can then be used to recover part of the heat of the initial condensate.
After use of flash steam the condensed water has to be reheated to its initial saturated pressure
and temperature for regeneration to high(er) pressure steam. As a result, the energy balance of
the total system is equal with or without the use of flash steam (assuming no extra heat losses).
As a rule of thumb, returned condensate after the use of flash steam has a temperature of 80 oC.
(US Department of Energy, 2006) At atmospheric pressure, h0 then is 335 kJ/kg.
81
Table 21: Steam tables
Enthalpy, kJ/kg Enthalpy, kJ/kg Enthalpy, kJ/kg
Temp,
T oC
Sat
press,
Psat kPa
Sat
liquid,
hf
Evap.,
hfg
Sat
vapor,
hg
Temp,
T oC
Sat
press,
Psat kPa
Sat
liquid,
hf
Evap.,
hfg
Sat
vapor,
hg
Temp,
T oC
Sat
press,
Psat kPa
Sat
liquid,
hf
Evap.,
hfg
Sat
vapor,
hg
0.01 0.6117 0.001 2500.9 2500.9 130 270.28 546.38 2173.7 2720.1 260 4692.3 1134.8 1661.8 2796.6
5 0.8725 21.02 2489.1 2510.1 135 313.22 567.75 2159.1 2726.9 265 5085.3 1159.8 1633.7 2793.5
10 1.2281 42.022 2477.2 2519.2 140 361.53 589.16 2144.3 2733.5 270 5503.0 1185.1 1604.6 2789.7
15 1.7057 62.982 2465.4 2528.4 145 415.68 610.64 2129.2 2739.8 275 5946.4 1210.7 1574.5 2785.2
20 2.3392 83.915 2453.5 2537.4 150 476.16 632.18 2113.8 2746.0 280 6416.6 1236.7 1543.2 2779.9
25 3.1698 104.83 2441.7 2546.5 155 543.49 653.79 2098.0 2751.8 285 6914.6 1263.1 1510.7 2773.8
30 4.2469 125.74 2429.8 2555.5 160 618.23 675.47 2082.0 2757.5 290 7441.8 1289.8 1476.9 2766.7
35 5.6291 146.64 2417.9 2564.5 165 700.93 697.24 2065.6 2762.8 295 7999.0 1317.1 1441.6 2758.7
40 7.3851 167.53 2406.0 2573.5 170 792.18 719.08 2048.8 2767.9 300 8587.9 1344.8 1404.8 2749.6
45 9.5953 188.44 2394.0 2582.4 175 892.60 741.02 2031.7 2772.7 305 9209.4 1373.1 1366.3 2739.4
50 12.352 209.34 2382.0 2591.3 180 1002.8 763.05 2014.2 2777.3 310 9865.0 1402.0 1325.9 2727.9
55 15.763 230.26 2369.8 2600.1 185 1123.5 785.19 1996.2 2781.4 315 10556 1431.6 1283.4 2715.0
60 19.947 251.18 2357.7 2608.9 190 1255.2 807.43 1977.9 2785.3 320 11284 1462.0 1238.5 2700.5
65 25.043 272.12 2345.4 2617.5 195 1398.8 829.78 1959.0 2788.8 325 12051 1493.4 1191.0 2684.4
70 31.202 293.07 2333.0 2626.1 200 1554.9 852.26 1939.8 2792.1 330 12858 1525.8 1140.3 2666.1
75 38.597 314.03 2320.6 2634.6 205 1724.3 874.87 1920.0 2794.9 335 13707 1559.4 1086.0 2645.4
80 47.416 335.02 2308.0 2643.0 210 1907.7 897.61 1899.7 2797.3 340 14601 1594.6 1027.4 2622.0
85 57.868 356.02 2295.3 2651.3 215 2105.9 920.50 1878.8 2799.3 345 15541 1631.7 963.4 2595.1
90 70.183 377.04 2282.5 2659.5 220 2319.6 943.55 1857.4 2801.0 350 16529 1671.2 892.7 2563.9
95 84.609 398.09 2269.6 2667.7 225 2549.7 966.76 1835.4 2802.2 355 17570 1714.0 812.9 2526.9
100 101.42 419.17 2256.4 2675.6 230 2797.1 990.14 1812.8 2802.9 360 18666 1761.5 720.1 2481.6
105 120.9 440.28 2243.1 2683.4 235 3062.6 1013.7 1789.5 2803.2 365 19822 1817.2 605.5 2422.7
110 143.38 461.42 2229.7 2691.1 240 3347.0 1037.5 1765.5 2803.0 370 21044 1891.2 443.1 2334.3
115 169.18 482.59 2216.0 2698.6 245 3651.2 1061.5 1740.8 2802.3 373.95 22064 2084.3 0.0 2084.3
120 198.67 503.81 2202.1 2705.9 250 3976.2 1085.7 1715.3 2801.0
125 232.23 525.07 2188.1 2713.2 255 4322.9 1110.1 1689.0 2799.1
82
User and database input
Input values required from user DATAPHYSCHEM DATABUO
Amount of steam
Temperature and saturated
pressure of the steam
Temperature and pressure of
water arriving at the boiler
Enthalpy of evaporation
Enthalpy of saturated water
Enthalpy of water arriving at
boiler
Boiler efficiency
Distribution and heat transfer
efficiency
User output
Amount of fuel needed for steam production
Available energy for heating processes
Impacts from steam generation (resulting from burning the fuel)
UPR output
Required amount of fuel, e.g.:
From technosphere:
Natural gas, burned in industrial furnace >100 kW, RER
Light fuel oil, burned in industrial furnace 1 MW, non modulating, RER
Example
For a certain process 1 tonne of steam is needed with a temperature of 165 oC. A closed system is
assumed. The efficiency of steam distribution and heat transfer is 92.5%. The boiler efficiency is
89%. The evaporation enthalpy of this type of steam is 2.0656 GJ/t. This means that 2.51 GJ of
fuel are needed to deliver 1 tonne of steam (useful heat).
Addendum: Analysis of natural gas
Natural gas can be used for delivering the heat. However, when calculating environmental
impacts from the use of natural gas it appears that different datasets report completely different
primary energy requirements. Therefore an analysis has been made of the datasets that are
incorporated in the Ecoinvent process ‘Natural gas, burned in industrial furnace >100 kW, RER’ in
Box 1 below.
If the Ecoinvent dataset ‘Natural gas, burned in industrial furnace >100 kW, RER’ is used, the ERE
(Energy Requirements for Energy) is 1.19 MJp/MJ. However, several different datasets have been
used as in input to this dataset. As can be seen, the country of origin determines to a large
extend the ERE with gas from Russia having the highest ERE, while gas from the Netherlands has
lowest ERE (1.07 MJp/MJ). Including the ERE in the calculation above (2.51 GJ/t steam) results in
a total non-renewable energy use (from cradle) of 2.99 GJ per tonne of steam (2.99 = 2.51 *
1.19).
83
Box : Analysis of fossil energy requirements of Ecoinvent dataset “Natural gas, burned in industrial furnace >100 kW/RER”
Amount Fossil energy
Natural gas, burned in industrial furnace 1 MJ 1.19 MJ
Natural gas, high pressure, at consumer 1 MJ 1.18 MJ
Natural gas, at long distance pipeline 0.0272 m3
1.16 MJ
m3
m3
m3
Natural gas, prod DE 0.05 x 0.0272 = 0.0014 0.055 MJ
Natural gas, prod DZ 0.16 x 0.0272 = 0.0044 0.190 MJ
Natural gas, prod GB 0.04 x 0.0272 = 0.0011 0.044 MJ
Natural gas, prod NL 0.24 x 0.0272 = 0.0065 0.257 MJ
Natural gas, prod NO 0.17 x 0.0272 = 0.0046 0.187 MJ
Natural gas, prod RU 0.34 x 0.0272 = 0.0092 0.432 MJ
---------------
1.165 MJ
MJ/m3
m3
Natural gas, prod DE 40.5 x 0.0272 = 1.10 MJ
Natural gas, prod DZ 43.65 x 0.0272 = 1.19 MJ
Natural gas, prod GB 40.6 x 0.0272 = 1.10 MJ Remaining fossil energy for:
Natural gas, prod NL 39.4 x 0.0272 = 1.07 MJ
Natural gas, prod NO 40.4 x 0.0272 = 1.10 MJ Process energy
Natural gas, prod RU 46.7 x 0.0272 = 1.27 MJ Transport
m3
Gas, natural, in ground 0.0272 1.04 MJ
84
2.5 Processing and use of materials
2.5.1 Electric manufacturing processes
To quantify electricity requirements for electric and electronic devices used in manufacturing is
not straightforward; many types of equipment perform completely different functions.
Nevertheless some generic relationships can be defined. In all manufacturing processes, the
energy requirement for the process can be divided into two components (Branham & Gutowski,
2010):
the base (auxiliary) power required to run all supporting process operations
the variable power needed to effect the physical transformation associated with the process.
Branham and Gutowski (Branham & Gutowski, 2010) illustrate this with the example of metal
casting; In metal casting, the base power would correspond to that required to heat the crucible
and operate mechanical components of the system. The variable power would correspond to the
incremental additional power required to heat and melt a unit of metal.
In this basic engineering module, we will estimate the energy use of electric manufacturing
equipment.
System boundary
The system boundary of this BUO includes the electric manufacturing equipment. This machine
might apply several processes, all of which are covered in the system boundary by the ‘electric
equipment’. We do not assume any additional processes in the system boundary.
Calculation algorithm
As a first approximation, the total power consumption rate (P in kW) can be modeled as follows:
vkPP 0
In which:
P = Total power (W)
P0 = Idle / auxiliary power (W)
k = Constant (J/unit processed)
v = Rate of material processing (unit processed/s)
The idle / auxiliary power represents the amount of electricity needed to start-up and maintain
the equipment in the ‘ready position’. The constant k represents the extra energy required to
process the material and could be for example melting energy, cutting energy, heating energy
etcetera. It is specific to the process and type of material and depends on properties such as
Electric
equipment
System
boundary
System
boundary
Electricity (En)
Input materials
(M)
Processed materials
(UO)
85
material hardness, heat capacity, temperature and enthalpies of any phase changes that might
take place.
Dividing both sides of the above equation by the throughput v yields an expression for the
specific energy consumption (SEC) of a process:
kv
P
v
PEelectr
0
In which:
Eelectr = Specific electrical energy per unit of material processed (J/unit)
User and database input
Input values required from user DATAPHYSCHEM DATABUO
Idle / auxiliary power
Rate of material processing - Constant of material conversion
User output
-
UPR output
From technosphere:
Amount of electricity required, e.g. “Electricity, medium voltage, production UCTE, at grid”.
Example
Chemical vapor deposition occurs with equipment with a power of 16 kW. No additional energy
requirements represented by k are needed. The throughput is 15 wafers/h (Murphy et al., 2003).
The energy use per wafer, hence, is 3.84 MJ / wafer.
Rule of thumb values
Several authors have estimated energy use of different types of equipment that could serve as a
rule of thumb. They are listed below. It should however be noted that energy use of electric
equipment completely depends on the type of equipment, the size, the brand and so on. It is
therefore impossible to provide rule of thumb values for a single process such as ‘injection
molding’. Nevertheless, the tables below (Table 22-26) give some indication.
Table 22: Energy use of different types of injection molding machines (Thiriez, 2005)
Injection molder type Material Power draw
(kW)
Throughput
(kg/hr) SEC (MJ/kg)
Parallel circuit PS 55.6 45.35 4.41
Sequential circuit PS 38.9 47.62 2.94
Fixed pump Nylon 6.7 10.31 2.35
Variable pump Nylon 3.55 10.69 1.20
Fixed pump ABS 23.4 31.75 2.65
Variable pump ABS 20.0 32.20 2.24
Servo pump ABS 11.5 32.47 1.27
Electric machine ABS 5.7 32.29 0.64
Variable volume pump Surlyn 18.3 27.57 2.39
Variable speed drive on
motor Surlyn 13.7 27.57 1.79
Electric Surlyn 6.3 28.53 0.80
Hydraulic PP 97.6 151.93 2.31
86
Injection molder type Material Power draw
(kW)
Throughput
(kg/hr) SEC (MJ/kg)
Hydraulic PP 65.7 99.32 2.38
All electric PP 51.4 154.20 1.20
Hydraulic TPP black 26.07 4.48 20.94
All electric TPP black 5.71 4.48 4.59
All electric TPP black 5.96 5.04 4.25
Full hydraulic HDPE 130.6 188.66 2.49
Electric screw drive HDPE 84.2 180.5 1.68
All electric HDPE 63.6 213.15 1.07
87
Table 23:Energy use of four different types of milling machines (Dahmus & Gutowski, 2004)
88
Table 24: Energy use for the production of a 0.13-µm metal microprocessor. The active area on the wafers is 261 cm2 consisting of 1 cm2 dies (Murphy et al., 2003)
Unit operation Throughput
(8-in. Wafers/h)
Power (kW)
Process Idle
Implant 20 27 15
Chemical vapor
deposition (CVD) 15 16 14
Wafer clean 150 8 7.5
Furnace 35 21 16
Furnace (rapid thermal
processor) 10 48 45
Photo (stepper) 60 115 48
Photo (coater) 60 90 37
Etch (pattern) 35 135 30
Etch (Ash) 20 1 0.8
Metallization 25 150 83
Chemical mechanical
polishing (CMP) 25 29 8
Table 25: Ductile iron induction melter energy usage per tonne shipped (Jones, 2007)
Process step Percentage of energy use (%) Energy use
(MJ/tonne shipped)
Melting 55 16079.24
Molding 12 3508.2
Core making 8 2338.8
Ladle transfer 4.5 1315.57
Heat treatment 6 1754.1
Finishing 7 2046.45
Other 7.5 2192.62
Total 100 29234.98
Table 26: Energy use of two different waterjet machines (Kurd, 2004)
Machine Power (kW) Cutting time
(min) Energy use (kWh)
Volume of material
removed (cm3)
OMAX-2652 12.25 1.21 0.247 0.98
OMAX-2626 4.87 1.416 0.115 0.79
Material requirements as a function of mechanical properties
For a given application, designers often can choose between numerous possible materials. The choice
is made based on material specific properties, such as costs, strength, weight, fire resistance,
tolerance to abrasion, permeability of gases etcetera. Even when a choice for a specific material has
already been made, new materials will often be developed to replace the initially chosen materials.
An example is the use of nanocomposites. Nanocomposites are polymers that are reinforced with
nanoobjects (i.e. particles with at least one dimension in the range of 1-100 nm) and could replace
conventional polymers or even steel or aluminium for a given application. If the material properties
of the novel material are different compared to the conventional material, the amount of material
needed will in many cases also differ. If a novel material is stronger, for example, less material will
be needed for the same function compared to a conventional material. Although the exact material
requirements should be determined with extensive technical material and product analyses, Ashby
(Ashby, 2005) defined material indices that could be used for a first estimate of material
requirements of a novel material compared to a conventional material. In this basic engineering
89
module, we will present the approach for determining material requirements based on the Ashby
material indices.
System boundary
The α system boundary for this BUO includes a structural application for which a certain amount of
material A is needed. This material A can be replaced by another material B to perform the same
function. The production of the material is included in the β system boundary and hence will not be
taken into account in this BUO. The amount of alternative material B for the same structural
application is the useful output (UO) of this BUO.
Calculation algorithm
The Ashby method for determining material requirements is represented by the following equation:
2
112
k
kmm
In which
m1 = Material requirement material A (kg)
m2 = Material requirement material B (kg)
k1 = Ashby index of material A (unit depending on index chosen)
k2 = Ashby index of material B (unit depending on index chosen)
Ashby’s material index (k) is a function of material properties such as the Young modulus (E), tensile
strength σf) and density (ρ). Ashby developed equations for different functional requirements, such as
strength or stiffness. It is assumed that material requirements are directly (inversely) proportional to
the material indices. The larger the value for k is, the better the mechanical properties of the
product under consideration are and the less material is needed for a certain function. The material
indices developed by Ashby are based on design that is ‘stiffness-limited’ (the material should not
bend), ‘strength-limited’ (the material should not break), ‘vibration-limited’ (the material should be
tolerant to vibration), ‘damage-tolerant’ (the material should resist damage), ‘electro-mechanical’
(the material is designed for electrical operation) or ‘thermal and thermo-mechanical’ (the material
should resist heat). Within each category, the appropriate function and constraints are chosen
(functions could be e.g. ‘tie’, ‘shaft’, ‘beam’, ‘column’, ‘panel’ or ‘plate’; constraints could be e.g.
required specification of dimensions). Tables B1-B7 show the various Ashby indices for different
functional requirements.
Structural
application
System boundary
System boundary
Material A (M)
Structural
application
System boundary
System boundary
Material B (UO)
90
91
92
93
94
95
User and database input
Input values required from user DATAPHYSCHEM DATABUO
Material requirement conventional
material
Specify function and constraint
Young modulus
Density
Failure strength
Shear modulus
Energy content/kg
Hardness
Damping coefficient
Fracture toughness
Thermal conductivity
Thermal diffusivity
Specific heat capacity
Thermal expansion coefficient
Electrical resistivity
Yield strength
Endurance limit
Material cost/kg
Ashby index
User output
Material requirement of alternative material for a specified function/application (kg)
UPR output
From technosphere:
kg material needed
Example
The following situation is considered: for the covering of a tomato greenhouse, plastic sheet is used.
Normally, such a sheet is made from low-density polyethylene (LDPE). However, a novel material has
been developed consisting of polypropylene reinforced with 3 wt% nanoobjects, i.e. organophilic
montmorillonite (‘polypropylene nanocomposite’). The question is what the material requirements
are for the novel material compared to the conventional LDPE.
It is assumed that in this case, the application is “strength-limited” (the material should not tear
apart, but it is allowed to bend) and the function is ‘tie’ (the forces working on the sheet are parallel
to the surface of the sheet). For this situation, the appropriate Ashby index is:
fk
The total amount of LDPE required to cover a standard greenhouse (k1) with a surface area of 430 m2
is estimated to be 2.38 tonnes. Furthermore, the following values apply:
σf of LDPE = 24 MPa
ρ of LDPE = 0.923 t/m3
σf of PP-nanocomposite = 37.4 MPa
ρ of PP-nanocomposite = 0.928 t/m3
Applying the basic engineering module, the material requirements for the tomato greenhouse using
the novel polypropylene nanocomposite are 1.51 tonnes.
2.5.2 Fuel use of cars
In car design and manufacturing, the ultimate goal is to develop cars that use as little fuel as
possible. One of the options to reduce fuel use is bringing down the weight of the car by using
96
lightweight materials. Fuel use is to a certain extent weight-dependent. In this basic engineering
module, we will present the relationship between the fuel use of a car and its weight. When the
weight of a car is known as well as a ‘new’ weight resulting from lightweight materials, its fuel
consumption can be estimated.
System boundary
The system boundary for this basic engineering module includes the performance of a car with a
certain weight 1 and a car with an alternative weight 2. In other words, it comprises the amount of
driven kilometers at a certain amount of fuel consumed. The distinction between cars with two
different weights is made because as we will see in the next section, for one calculation algorithm
the fuel use of a car with weight 2 is related to the fuel use of a car with reference weight 1. There
are no extra processes included in the β system boundary.
Calculation algorithm
Two sources have reported formulas for the calculation of fuel use of a car with a given weight. These
are Koffler and Rohde-Brandenburger (Koffler & Rohde-Brandenburger, 2010) and Maclean and Lave
(Maclean & Lave, 2000). We will first present the algorithm by Koffler and Rohde-Brandenburger.
Algorithm according to Koffler and Rohde-Brandenburger
In order to quantify the fuel consumption of any vehicle, the underlying driving cycle has to be
specified first. The driving cycle specifies speed and chosen gear as a function of time. In Europe the
driving cycle has been standardized in the New European Driving Cycle (NEDC). It consists of 780
seconds (s) in an urban cycle and 400 s in an extra-urban cycle. Its average speed is 33.6 km/h, the
maximum speed is 120 km/h and the total distance is 11 km. Figure 11 illustrates the NEDC.
Driving car with
weight 1
System boundary
System boundary
Fuel
(M/UO)
Kilometers
driven (UO)
Driving car
with weight 2 Fuel
(M/UO)
Kilometers
driven (UO)
97
Figure 11: Time and speed pattern of the New European Driving Cycle
Following the NEDC profile, a vehicle’s power train has to provide the traction force to overcome
three types of driving resistance:
Rolling resistance (in N): RR fgmF
Aerodynamic resistance (in N): AvFL
2
2
Acceleration resistance (in N): amFa
In which:
m = Vehicle weight (kg)
g = Gravitation constant (m s-2)
fR = Rolling resistance coefficient (dimensionless)
ρ = Air density (kg m-3)
v = velocity (m/s)
α = Air drag coefficient (dimensionless)
A = Front surface (m2)
The mechanical work W (J) can be obtained through their respective integration over the distance s
(m):
dsFW
Since the NEDC’s velocity profile cannot be expressed through a simple mathematical function, the
integral is calculated as the sum of all work increments in between t=0 and t=1180 s.
WRRR CfgmW
WLL CAW2
Waa CmW
98
The characteristic values CWR, CWL and CWa are constants that are independent of the respective
vehicle and specific to the driving cycle. Table 27 shows values for the European, the US and the
Japanese driving cycles.
Table 27: Characteristic values CWR, CWL and CWa for different driving cycles
NEDC (EU) Combined fuel
economy (US) 10-15 mode (JP)
CWR (m) 11,013 17,198 4,165
CWL (m3/s2) 3,989,639 6,341,415 699,767
CWa (m2/s2) 1,227 2,221 687
About 15% of the NEDC total distance of 11 km is deceleration phase. During deceleration, no energy
is needed. Because rolling resistance is virtually independent of the vehicle’s velocity, the related
work also decreases by 15%. The total energy demand becomes:
aLRsum WWWW 85.0
The degree of efficiency η of an internal combustion engine (ICE) depends to a large extent on its
point of operation concerning revolutions per minute (rpm) and engine output. In general it can be
characterized as follows:
totalin
totalout
totalP
P
,
,
In which:
Pin = Energy input in liters fuel per hour (l/h)=(kW)
Pout = Engine output in kilowatts (kW)
The unit could be expressed as l h-1kW-1, which equals to l/kWh or l/MJ.
The fuel consumption characteristics of an ICE can be depicted with the help of the Willans line
method (Ross, 1997). Especially for low output and low rpm (<4,000 min-1), which are typical for the
NEDC, the Willans lines run almost parallel. This is shown in Figure 12:
99
Figure 12: Willans lines and resulting trend lines of a 1.4 l TSI gasoline engine (90 kW) for low output and low rpm
If it is further assumed that 2% of the engine’s additional energy output is lost in the gearbox, fuel
consumption calculates to:
02.1sumNEDC WV
In which:
VNEDC = Fuel consumption in the NEDC (l/11 km)
Algorithm according to Maclean and Lave
Maclean and Lave (Maclean & Lave, 2000) have estimated the fuel use of a car empirically based on
fuel consumption of a car with reference weight 1. The relationship they have derived is represented
by the following formula:
72.0
2
112
W
Wff
In which:
f1 = Efficiency of car with reference weight 1 (km/l)
f2 = Efficiency of car with new weight 2 (km/l)
W1 = Reference weight 1 of car (kg)
W2 = New weight 2 of car (kg)
Comparison of the two algorithms
Both algorithms just outlined (section 3.2.1 and 3.2.2) can be used to calculate the car
efficiency/fuel use of cars at different weights. If car efficiency is plotted against car weight at
different car weights and a reference car efficiency of 20.4 km/l at a weight of 2000 kg, the lines
coincide rather well as is shown in Figure 13, although the lines start deviating more at lower car
weights.
100
Figure 13: Comparison of car efficiency results using the two available algorithms
User and database input
Input values required from user DATAPHYSCHEM DATABUO
Vehicle weight
Front surface area of car
Air density
Air drag coefficient
Gravitation constant
Constant CWL
Constant CWR
Constant CWa
Engine efficiency
User output
Car efficiency / fuel consumption
UPR output
The amount of fuel consumed by the car for a certain distance, e.g. “Petrol, unleaded, at
regional storage”.
Example
We consider a Mercedes Benz W140, type ‘sedan’. This car has a weight of 2000 kg. Furthermore, the
following parameter values apply:
Front surface area: 2.8 m2
Air drag coefficient: 0.39
Engine efficiency: 0.076 l/MJ
Rolling resistance: 0.01
Air density: 1.225 kg/m3
Gravitation constant: 9.81 m/s2
CWL: 3989639 m3/s2
CWR: 11013 m
CWa: 1227 m2/s2
WL, WR and Wa can now be calculated:
WL = 2.67 MJ
WR = 2.16 MJ
15.0
17.0
19.0
21.0
23.0
25.0
27.0
29.0
31.0
1000 1500 2000 2500 3000 3500
Weight of car (kg)
Car
eff
icie
ncy (
km
/l)
Koffler and Rohde-
Brandenburger
Maclean
101
Wa = 2.45 MJ
Wsum amounts to 6.96 MJ and, hence, VNEDC is 0.54 l/11 km which equals to 20.4 km/l
It should be noted that the results for fuel use/car efficiency are quite optimistic when using the
constants for the NEDC. If constants for the combined fuel economy (US) are used, this car would
drive only 12.3 km/l. If the weight of this car would be reduced with 100 kg, the related fuel use can
be calculated with the formula from Maclean and Lave (Maclean & Lave, 2000):
f1 = 20.4 km/l
W1 = 2000 kg
W2 = 1900 kg
Hence, the new car efficiency ‘f2’ equals to 21.1 km/l.
102
Part II: Validation of the engineering modules
103
1. Introduction
In the transition to a more sustainable society, Life Cycle Assessment (LCA) is considered as a valuable
assessment framework for the quantification of the environmental impact of products and services
(European Commission, 2003). For this purpose it is more and more included in the decision making
process at different levels of policy and industry. The LCA supported decision should then make a
choice between options by outweighing the positive and negative environmental effects of each
option. Obtaining meaningful results requires that all options are thoroughly known; LCA is indeed a
data intensive procedure (Mueller et al., 2004) that typically requires full material and energy
balances of the system under study. However, this might not be available for the different options
under development, which is a major shortcoming for the implementation of prospective
environmental sustainability assessments.
Especially in biorefineries this is an important issue. The transition from fossil based to biobased
refineries is seen as a big step in greening the economy by mitigating climate change and offering
new options to obtain renewable and thus more sustainable products and services (IEA Bioenergy,
2009b). Nevertheless, it is already shown that this transition is challenging as renewable resources
also have a production chain with significant environmental impact, e.g. through land and water use
(De Meester et al., 2011). Strategic choices should therefore be made before final implementation in
order to achieve the highest degree of sustainability with the biomass that is available.
The main problem to assess future biorefineries is obtaining a data inventory with a full mass and
energy balance of the new processes. For this purpose in part I, basic engineering modules were
developed that can be used as parameterized unit operations where raw data and formulas are used
to determine the data inventory, instead of using fixed values only (Cooper et al., 2012). This allows
the completion of the life cycle inventory by assembling material and energy balances of basic unit
operations (BUO; single process steps such as a distillation column) whilst being flexible and
straightforward in use and easily accessible to life cycle practitioners as the modules can be
implemented in LCA software such as OpenLCA. In this part of this report, the modules developed are
validated in a biorefinery case study to demonstrate the operability and accuracy of the proposed
approach.
2. The biorefinery case study
The basic engineering approach is tested in a biorefinery case study in which wheat, flour and sugars
are processed into specialty sugars, starch, gluten, ethanol and animal feed (Figure 14). The
conversion of the biofeedstock concerns typical processes of a biorefinery, starting with dry and wet
milling steps and further separations mainly with centrifuges and sieves. Afterwards the intermediate
flows are upgraded to final products, mainly by drying and evaporation. In this type of biorefinery,
not many ‘reactions’ occur as it is more the separation of the different molecules of the biofeedstock
as such, possibly after a hydrolysis step. There are two exceptions, namely the fermentation of wet
starch streams to bioethanol with a subsequent distillation section, and a combined heat & power
(CHP) engine where natural gas is converted into steam, hot water and electricity.
104
Figure 14: A process diagram of the case study
The case study is a complex system with a large number of single processes where an extensive
dataset was gathered as a year average of 2009 for many of these processes. Yet, it was not possible
to collect all data of all processes. Therefore a selection has been made of ten of the twenty-two
processes limited by confidentiality, availability of detailed data and availability of a specific BUO
that is applicable for the considered process. This validation effort is still valuable as it gives a first
indication on the accuracy of the results and it can serve as guidance for further work on the
quantification of life cycle inventories for prospective assessments by means of parameterized
modules. In total, ten of the twenty-two BUO approaches were tested, often based on several
subsamples within their own α boundary. This test is elaborated in Table 28, where also the sample
size is given. In total, 41 subsamples were tested, where identical processes (e.g. two centrifuges)
are only counted once.
105
Table 28:The elaborated BUO categories with case specific information and sample size
BUO Case specific process information Sample size
Comminuting The wheat grains are first dry-milled by using a complex
system of rolls and rotors
3
Agitation and mixing of
liquids and suspensions
Both the agitation in the fermentation tanks and the dough
mixer are studied
3
Sedimenting centrifuges Centrifuges are mainly used for wet separation of dough,
gluten, etc.
7
Evaporation After wet separation, the dry matter content of flows is
increased by evaporators with 1 or more effects.
3
Mechanical compression –
single stage
Recompression is used to upgrade vapor of evaporators 4
Fermentation Wet starch streams of the factory that are less suitable as
food or feed are fermented to bioethanol
1
Binary distillation The ‘beer’ solution obtained from fermentation is distilled to
purify the ethanol
1
Heating Several flows are preheated before evaporation, drying, etc. 5
Incineration for heat and
power
A CHP working on natural gas produces hot water, steam and
electricity to supply the factory with energy. Excess
electricity is sold to the local grid
1
Conveying solids Before the wet milling steps, biomass is mainly transported
and ensilaged by different types of conveyors
13
The results of the modeling of mass and energy balances within the α system boundary are validated
with the actual mass and energy balances of the factory. As the sample size (n) was limited for some
operations, no statistical uncertainty is elaborated. Instead, per calculation of a basic unit operation
within the α system boundary, an Accuracy Factor and a Relative Approximation Error (RAE) is
calculated, which are defined respectively as:
In this validation procedure, the products of the biorefinery are assessed in a life cycle perspective.
The parameterized BUO modules are used mainly to predict energy consumption of the different
processes. The accuracy of this calculation is then tested with the actual data obtained from the
factory. Afterwards, the results are linked to life cycle assessment by calculating a Carbon Footprint
based on the IPCC 2007 (Solomon et al., 2007) impact assessment method, relying on the ecoinvent
database for the datasets to model the γ system boundary.
106
3. Results and discussion
Within the aforementioned limitation, it was possible to calculate 27% of heat and 42% of the
electricity use of the factory. In the following, first two examples of BUO validations are presented.
Second, an overall overview on the tests is discussed and third, the application in a Life Cycle
Assessment is analyzed.
3.1 Validation examples
3.1.1 Mechanical compression – single stage
Most compressors in the factory are used for recompression of vapour in evaporator systems. One of
the compressors upgrades the vapour of 377K to usable steam of 439K and has an actual energy use of
2599kJ/kmol. Using the calculation procedure for this BUO with the values given in yields a predicted
energy use of 2829 kJ/mol (Table IV.1.7), obtaining a Relative Approximation Error of 9% for this
process. By applying this BUO approach, the output temperature and pressure are obtained, allowing
further calculations with the available heat of the steam, and the electricity use of the compressor is
known when linked to the flow rate. Coupling this to a life cycle database allows obtaining the carbon
footprint of the operation. Based on the Belgian electricity mix (medium voltage), the carbon
footprint of the compression of 100 kg steam according to the parameters in Table 29Fout!
Verwijzingsbron niet gevonden. results in an emission of 1.48 kg CO2-eq according to the
calculations, whereas the actual emission is 1.36 kg CO2-eq.
Table 29:Parameters in the calculation procedure of the BUO mechanical compression – single stage
DATA source Symbol Value Unit
INPUT pi 101325 Pa
INPUT Ti 377 K
INPUT Tc 439 K
DATAPHYSCHEM R 8.31 J/mol*K
DATAPHYSCHEM γ 1.31
DATABUO α 77 %
OUTPUT Pc 192769 Pa
OUTPUT pc/pi 1.90
OUTPUT W 2829 kJ/kmol
3.1.2 Evaporation
After the different wet milling and separation steps, different starch streams require higher dry
matter concentrations. One of the streams enters in an evaporator at a dry matter concentration of
2.8% and leaves at 4.9%. The actual energy delivered by the steam is 998kJ per kg incoming feed,
whereas the calculated value indicates 973kJ per kg (Table 30), resulting in a RAE of 3%. Using this
BUO thus allows to estimate the heat requirement of the evaporation process. Coupling this to a life
107
cycle database gives a Carbon Footprint of 2.65 kg CO2-eq according to the calculations, and 2.72 kg
CO2-eq based on the actual data (by using Heat, at cogen 1MWe lean burn, allocation exergy/RER).
Table 30:Parameters in the calculation procedure of the BUO evaporation
DATA source Symbol Value Unit
INPUT mf 1.00 kg
INPUT mo 0.56 kg
INPUT Ti 108.0 K
INPUT Tb 108.6 K
DATAPHYSCHEM cp 4.10 kJ/kg*K
DATAPHYSCHEM λ 2230 kJ/kg
DATABUO η 90 %
OUTPUT q 973 kJ
3.2 Overview of the validation
Figure 15 summarizes the Accuracy Factors obtained by the validation of the different samples, where
1 would be a perfect fit of the actual mass and energy balance of the factory. In this validation test,
an overall average relative approximation error of 22% is obtained.
108
Figure 15: A summary of the Accuracy Factors of the different BUO calculations. A value of 1 represents a perfect fit of the calculated result over the real value
The Combined Heat and Power (CHP) plant is the starting point and is accurately predicted with a
RAE of 0% for electricity and 2% for heat production. The standard efficiency factors for an efficient
natural gas-fired CHP producing electricity and heat are thus reliable in this case, which is essential
in the sustainability assessment of the factory as this CHP delivers energy to all other operations and
thus has a large influence on the final result. It should, however, be noted that the heat availability is
predicted accurately, but not necessarily all heat is required in the processes. The allocation factors
in LCA studies should thus be based on the real heat use instead of the potential heat use.
The calculations for milling (comminuting) result in a RAE of 7%. A good Bond Work Index should,
however, be found for the milled material, which might not be available in some cases. For wheat
grains however, a reliable value was found in literature (Das, 2005). It should also be noted that the
calculations are valid if the power of the mill is put directly on the particle. In case the particle size
is reduced by rotational forces causing friction between the particles, other calculation procedures
should be sought.
The deviation in the agitation model is larger; the validation highlights RAE values of up to 107% for
the fermentation and 27% for the dough mixer. As a clarification the used equations in the
parameterized module of agitation power (P) should be considered (McCabe et al., 2004):
for the turbulent region and
for the laminar region. With µ the viscosity, ρ the density and Np and KL flow regime dependent pump
numbers. The prediction of equipment power thus depends heavily on the impeller velocity (N) and
diameter (Da). As these parameters are in most cases approximated, small deviations from the real
value result in a large sensitivity of the final result. Furthermore, this calculation requires fluid
109
properties, which are approximated by the Power Law for ‘slurry’ type of flows, resulting in an
additional potential source of deviation. The same is true for pumping incompressible fluids in a
biorefinery. It is very difficult to obtain all exact parameters for the calculation (e.g. amount of
turns, length of pipes, etc.). This is more straightforward for conveyors where the module basically
requires only the type of conveyor, the flow rate and an approximation of the length. The calculations
based on 13 samples yield RAE of 23%, which is reasonable as a first estimate, knowing that this
operation is of a smaller relevance relative to the total power use in the factory.
The average RAE for sedimenting centrifuges is 25% if each type is counted once. This factor is lower
for the total absolute centrifuge power (10%) as the most accurately predicted centrifuges have the
largest contribution to total power use. In 4 of 7 samples the power was predicted accurately
(average RAE = 8%), but in the other samples a larger RAE was obtained (33 to 60%). This is the
disadvantage of the fact that this module is not based on a continuous equation, but on integer
ranges depending on typical combinations of liquid and solid throughput. The largest deviation occurs
when it is not obvious to which exact range the centrifuge should be assigned.
The fermentation section was predicted with a RAE of 3% for the fermentation yield and 4% for the
energy use of the distillation column. This module is however highly sensitive to the reflux ratio
applied in the column, as this is multiplied with the latent heat, causing the largest part of the
energy requirement. The (pre)heating steps are predicted with a relative approximation error of 4%
and also the heat use per evaporator is predicted fairly accurate (RAE = 8%) based on the
thermodynamic balance. The configuration of the different evaporators can however be complex,
with different effects and with intermediate compressors to upgrade vapor streams. As a result, the
total heat required is not the sum of each heat requirement separately. Instead the BUO modules for
heat consumption also require heat integration and consideration of valorization of ‘waste’ heat from
other processes and of the use of recompressors to upgrade vapor. The compressors themselves are
predicted relatively accurately (RAE = 6%). It should however be noted that two adjustments had to
be made; first one of the compressors is known to have a lower efficiency than ‘traditional’ current
equipment where a default value of 77% is assumed for centrifugal compressors. Second, it is possible
that in industry equipment is reused from a previous application and therefore not optimally sized for
their current task, which might lead to additional energy consumption, but which is sound from an
economic perspective.
3.3 Using the BUO approach in LCA
The carbon footprint (γ system) of the calculated modules of the factory is 96% of the carbon
footprint based on actual data when excluding the ‘milling’ steps based on rotational forces causing
friction and including only the optimally designed compressors (87% including these excluded
processes). This good score can be explained by the fact that the results of the parameterized
modules with the highest accuracy factor have a large share in the absolute carbon footprint of the
factory (Figure 16). Additionally, the obtained impact is the sum of energy use in the α system
boundary of the ten applied BUO approaches and has a better score than each BUO separately, which
can be explained by considering Figure 5, from which no bias of over- or underestimation can be
observed resulting in a compensation of higher and lower approximations. Furthermore, when
assembling the results of each BUO separately, we performed a similar heat integration over the
different modules compared to the real factory. This aspect requires care, as the sum of heat
calculated per module is much higher than real total heat use. The degree of heat integration
however, cannot be generalized as it depends on the specific process chain, availability of ‘waste
heat’ and company policy. Another difficulty is integration of the β system boundary in the
assessment. In the studied biorefinery these SUO operations are not negligible: approximately 22% of
total electricity use is used for pumping liquid fluids and as the biorefinery is situated in a densely
populated area, an elaborated system of controlling air streams by ventilation and aspiration is used.
This SUO system accounts for approximately 16% of the plant’s electricity use and was also not
validated because of its complexity. This system of SUO is difficult to generalize or to model
separately; information such as length and amount of turns in piping systems or the pressure drop in
ventilators is essential but needs detailed designing which is not straightforward. Another
110
shortcoming is the unavailability of the BUO drying, which accounts for almost all heat use that was
not incorporated in this validation. It is indeed a challenging task to find a suitable generic
engineering module for this process, as there is a large range of dryer types each with their specific
configuration. The energy demand of this process depends on this configuration and is a combination
of interlinked factors such as air humidity, temperature, velocity, surface area, equilibrium moisture
content, etc. A large range of company specific choices is possible impeding generic calculations,
such as the balance between more electricity to achieve a higher air flow rate over the material or
using hotter air to increase efficiency per volume of air used. On top of the parameterization of
energy consumption, also the mass flows in a studied system should be studied, as the impact of the
agriculture of the biofeedstock should also be allocated to the output products. In this work, the
fermentation yield was approximated with an relative approximation error of 3%. The other mass
flows are, apart from different hydrolysis steps, not chemical reactions, but are separations of the
different molecules of the plant (e.g. gluten, starch, …). The latter are user specific choices that are
rather straightforward and are therefore not included in this work. The use of utilities such as foam
inhibitors, silica, cleaning agents, buffers and enzymes is case specific and requires expert judgment
rather than generic parameterization. While the impact of the latter is relatively small
(approximately 6%), the combination of the different drawbacks might cause a significant increase in
the deviation of the total carbon footprint of a prospective LCA with the BUO approach.
Figure 16: The contribution of the tested modules to the final carbon footprint
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
Calculated values Real values
Conveying solids
Heating
Binary distillation
Mechanical compression – single stage
Evaporation
Sedimenting centrifuges
Agitation and mixing of liquids andsuspensions
Comminuting
111
General conclusions
The proposed approach of using engineering calculations, rules of thumbs and default values to obtain
the mass en energy balance of a system by applying a modular BUO approach is identified as a useful
and reliable way to support prospective life cycle assessments. Engineering calculations allow to
obtain UPR for other production chains than those currently available in literature and databases and
offer the freedom to be adjusted for the assessment of future technologies. The approach chosen can
be used to simulate datasets in the foreground system, obtaining data compatible with life cycle
practices, which allows coupling the modelled system to a background system. It is possible to work
with different levels of details of input data in the models. Default values are available in many cases
as approximations, but different parameters can be adjusted if more detailed data is available. This
type of work can become a new source of data for Life Cycle Assessments and can become an
invaluable help in prospective assessments. Furthermore, these modules can be implemented in LCA
software such as OpenLCA in order to become more user friendly.
In total twenty-five modules were developed in five categories:
Reactions
Separation processes
Physical mechanical processes
Utilities
Processing and use of materials
Of these twenty-five parameterized modules, ten were tested in the case study. Apart from agitation,
most of these modules yield relatively good results compared to the real values with an overall
average relative approximation error of 22%. When these modules are used in the case study to
estimate the carbon footprint of a biorefinery, an accuracy of 96% was achieved. It should, however,
be noted that there is a compensation between over- and underestimation in the case studies
conducted. Furthermore, three major shortcomings were identified. Firstly, whereas the inventory of
the α boundary was obtained with an acceptable error, the supporting unit operations such as
pumping are very specific and difficult to obtain whilst not being negligible. Secondly, a factory is
complex and processes are not independent from each other; a successful application of the BUO
approach therefore requires a careful integration of heat use instead of a linear summation of heat
use of different processes. Thirdly the list of selected BUO is still limited to twenty-two, whereas
other operations also occur in industry. For example in biorefineries, operations such as cyclones and
dryers are frequently used. More parameterized modules should thus be developed in the future.
Nevertheless, while more case studies are necessary for confirmation of the results in this study, the
currently available modules produce relatively reliable results. The possibility of integration of these
modules as parameterized unit operations in LCA software can be of major value to obtain a detailed
data inventory necessary for prospective sustainability assessments. The modules should, however, be
used with care and with a realistic design of a production chain in mind.
112
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